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		| Clement 
 
 
 Joined: 24 Apr 2006
 Posts: 1113
 Location: Dar es Salaam Tanzania
 
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				|  Posted: Wed May 29, 2013 10:23 pm    Post subject: May 30 VH |   |  
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				| X-Wing on 8 in r19c17; r8c1<>8, r2c7<>8 and 	  | Code: |  	  | +------------+----------+--------------+
 | *28  3  24  | 9   1 6  | *58   7   45  |
 | 1   68 5   | 2   4 7  | 689  6-89 3   |
 | 9   46 7   | 35  8 35 | 26   126 146 |
 +------------+----------+--------------+
 | 4   2  8   | 136 7 13 | 69   5   169 |
 | 6   5  3   | 14  9 2  | 7    14  8   |
 | 7   9  1   | 46  5 8  | 3    46  2   |
 +------------+----------+--------------+
 | 3   7  69  | 158 2 15 | 4    68-9 #569 |
 | 25-8 48 246 | 58  3 9  | 1    268 7   |
 | *258 1  29  | 7   6 4  | *2589 3   59  |
 +------------+----------+--------------+
 
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 XYZ-Wing 569 with apex in r7c9; r7c8<>9; stte.
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		| arkietech 
 
 
 Joined: 31 Jul 2008
 Posts: 1834
 Location: Northwest Arkansas USA
 
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				|  Posted: Wed May 29, 2013 11:21 pm    Post subject: Re: May 30 VH |   |  
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				|  	  | Clement wrote: |  	  | X-Wing on 8 in r19c17; r8c1<>8, r2c7<>8 and XYZ-Wing 569 with apex in r7c9; r7c8<>9; stte.
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 All i needed was the x-wing for lclste.
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		| Marty R. 
 
 
 Joined: 12 Feb 2006
 Posts: 5770
 Location: Rochester, NY, USA
 
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				|  Posted: Thu May 30, 2013 1:26 am    Post subject: Re: May 30 VH |   |  
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				|  	  | arkietech wrote: |  	  |  	  | Clement wrote: |  	  | X-Wing on 8 in r19c17; r8c1<>8, r2c7<>8 and XYZ-Wing 569 with apex in r7c9; r7c8<>9; stte.
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 All i needed was the x-wing for lclste.
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 The X-Wing did it for me as well.
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		| hughwill 
 
 
 Joined: 05 Apr 2010
 Posts: 424
 Location: Birmingham UK
 
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				|  Posted: Thu May 30, 2013 10:14 am    Post subject: Re: May 30 VH |   |  
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				| Looks to me as though the X wing on 8 and the XYZ on 569 are both single shots.....
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		| TerenceF 
 
 
 Joined: 21 Dec 2007
 Posts: 26
 Location: Takapuna, NZ
 
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				|  Posted: Mon Jun 03, 2013 4:59 am    Post subject: |   |  
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				| As an alternative, from the same starting point, there is an XY chain starting at r1c9 which eliminates the 5 from r7c9 thus: 
 If r1c9 = 5 then r7c9<>5 AND
 If r1c9 = 4 then r1c3=2 so r9c3=9 so r9c9=5 so r7c9<>5
 
 THis forces r7c8 = 8 which solves the puzzle.
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