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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Thu Jan 20, 2011 4:57 am Post subject: Puzzle 11/01/20 BBDB |
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Code: | +-----------------------+
| 6 . . | 4 2 . | 9 . 7 |
| . 2 . | 9 7 . | . . . |
| . . 4 | . 3 . | . 6 . |
|-------+-------+-------|
| 5 9 . | 2 6 4 | . . . |
| 7 8 6 | 5 . 9 | . . 4 |
| . . . | 3 8 . | . . 6 |
|-------+-------+-------|
| 8 . . | . . . | 2 . . |
| . . 2 | . . . | . . 9 |
| 4 . . | . 9 2 | . 8 3 |
+-----------------------+
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Play this puzzle online at the Daily Sudoku site |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Thu Jan 20, 2011 4:44 pm Post subject: |
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Four steps for me.
Quote: | XYZ-Wing (158)
X-Wing (1)
UR (45) with pseudo cell 17
W-Wing (67) |
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JC Van Hay
Joined: 13 Jun 2010 Posts: 494 Location: Charleroi, Belgium
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Posted: Thu Jan 20, 2011 6:59 pm Post subject: |
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One "simple"(!?) step proof ...
Quote: | From XY Ring in R39C47 : Either r3c4=1, NT(167)r379c4, r8c4=8, NP(13)r8c16, r8c2=6; Or r3c4=8, r3c7=1, r9c7=6 => r8c7<>6.
In Eureka notation : (6=1)r9c7-1r3c7=(167)r379c4-(167=8)r8c4-(8=136)r8c126 : 6r9c7=6r8c2 => -6r8c7.
8-SIS AIC : (61)R9C7 1R3 (1=67)R79C4 (167=8)R8C4 (8=13)R8C16 (13=6)R8C2 : 6r9c7=6r8c2 => -6r8c7. |
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peterj
Joined: 26 Mar 2010 Posts: 974 Location: London, UK
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Posted: Thu Jan 20, 2011 11:53 pm Post subject: |
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Another approach to make JC's elimination...
Code: | Kraken aahs (47)r8c78, (4)r8c5, (7)r8c4 ; r8c7<>6
hp(47)
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(4)r8c5 - (4=5)r7c5 - (5=1)r7c9 - (1=6)r9c7
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(7)r8c4 - anp(7=16)r79c4 - (1)r4c4=r4c7 - (1=6)r9c7 | ... but I actually played same moves as Marty's in a different order... |
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