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		| daj95376 
 
 
 Joined: 23 Aug 2008
 Posts: 3854
 
 
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				|  Posted: Sat Jun 19, 2010 3:19 pm    Post subject: Puzzle 10/06/19: (A) VH |   |  
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				|  	  | Code: |  	  | +-----------------------+ | 1 . . | . 4 6 | . 8 . |
 | . 5 . | 2 1 . | . . . |
 | . . 2 | . 3 . | . . . |
 |-------+-------+-------|
 | . 1 . | . . . | . . 6 |
 | 2 4 3 | . 6 1 | . 7 8 |
 | 6 . . | . 7 2 | . . . |
 |-------+-------+-------|
 | . . . | . . . | . 9 2 |
 | 4 . . | . 9 . | 6 3 . |
 | . . . | 1 2 . | 8 . 4 |
 +-----------------------+
 
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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: Sun Jun 20, 2010 2:44 pm    Post subject: |   |  
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				| After basics: Note the DP 79 in R149.  Either one of R4C13 is 5, and/or R1C2 is 3.  The 35 pseudocell forms an XY-wing 3-58, removing 5 in R6C3 and solving the puzzle. 	  | Code: |  	  | +-------------------+-------------------+-------------------+ | 1     379$  79    | 579   4     6     | 2     8     3579  |
 | 3789  5     4     | 2     1     789   | 39    6     379   |
 | 789   6     2     | 5789  3     5789  | 14    14    579   |
 +-------------------+-------------------+-------------------+
 | 579%  1     579%  | 34    58    589   | 34    2     6     |
 | 2     4     3     | 59    6     1     | 59    7     8     |
 | 6     89   -589   | 34    7     2     | 13459 14    39    |
 +-------------------+-------------------+-------------------+
 | 35    38@   1     | 6     58    4     | 7     9     2     |
 | 4     2     58#   | 578   9     578   | 6     3     1     |
 | 79    79    6     | 1     2     3     | 8     5     4     |
 +-------------------+-------------------+-------------------+
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 This is probably not what Danny had in mind ...
 
 Keith
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		| daj95376 
 
 
 Joined: 23 Aug 2008
 Posts: 3854
 
 
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				|  Posted: Sun Jun 20, 2010 4:31 pm    Post subject: |   |  
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				|  	  | keith wrote: |  	  | This is probably not what Danny had in mind ... 
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 Keith: Nice find!
 
 
  	  | Code: |  	  | <79> DP[(5)r4c13 = (3)r1c2]r149c123 - (3=8)r7c2 - (8=5)r8c3 => r6c3<>5 
 -or-
 
 (5)r6c3 - <79> DP[(5)r4c13 = (3)r1c2]r149c123 - (3=8)r7c2 - (8=5)r8c3 - (5)r6c3
 
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 What I had in mind is interesting as well.
 
 
  	  | Quote: |  	  | 2x concurrent UR Type 1 => r6c7<>134. 
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 Regards, Danny
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		| Marty R. 
 
 
 Joined: 12 Feb 2006
 Posts: 5770
 Location: Rochester, NY, USA
 
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				|  Posted: Mon Jun 21, 2010 9:54 pm    Post subject: |   |  
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				| Very nice Keith. 
 I used the overlapping Type 1 URs. After removing 134 from r6c7, the remaining 589 triple finishes things off.
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