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		| crunched 
 
 
 Joined: 05 Feb 2008
 Posts: 168
 
 
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				|  Posted: Thu Mar 05, 2009 2:58 am    Post subject: March 5, vh |   |  
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				| Here is the grid after basics. 
 
  	  | Code: |  	  | +---------------+-------------+-----------+
 | 9    247  1   | 3   6   27  | 5 28  48  |
 | 8    5    6   | 4   1   29  | 7 239 39  |
 | 27   247  3   | 5   8   279 | 6 1   49  |
 +---------------+-------------+-----------+
 | 2357 2379 257 | 279 259 4   | 8 6   1   |
 | 6    1    57  | 79  3   8   | 4 579 2   |
 | 4    279  8   | 6   259 1   | 3 579 59  |
 +---------------+-------------+-----------+
 | 125  8    25  | 12  7   3   | 9 4   6   |
 | 137  37   9   | 18  4   6   | 2 358 358 |
 | 23   6    4   | 289 29  5   | 1 38  7   |
 +---------------+-------------+-----------+
 
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 Play this puzzle online at the Daily Sudoku site
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		| Earl 
 
 
 Joined: 30 May 2007
 Posts: 677
 Location: Victoria, KS
 
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				|  Posted: Thu Mar 05, 2009 3:09 am    Post subject: Marach 5 VH |   |  
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				| The old reliable, an xy-wing (237), will solve it in one step. 
 Earl
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		| Clement 
 
 
 Joined: 24 Apr 2006
 Posts: 1113
 Location: Dar es Salaam Tanzania
 
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				|  Posted: Thu Mar 05, 2009 1:48 pm    Post subject: Daily Sudoku: Thu 5-Mar-2009 VH |   |  
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				| XY-Wing {237} Pivot {23}r9c1 eliminating 7's in r1c2 &r3c2 solves the puzzle. |  | 
	
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		| stevieboy 
 
 
 Joined: 25 Jan 2008
 Posts: 31
 Location: Michigan
 
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				|  Posted: Thu Mar 05, 2009 7:34 pm    Post subject: |   |  
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				| It seemed like it took years to find the (237) xy-wing, but, that's what broke it open! |  | 
	
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		| Fungii 
 
 
 Joined: 06 Mar 2009
 Posts: 2
 
 
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				|  Posted: Fri Mar 06, 2009 2:08 am    Post subject: |   |  
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				| I'm kind of new to these funky techniques you guys use to solve these puzzles, so I didn't see the (237) xy-wing.   
 Instead, I worked on the (5,7) r5c3. Both options worked to eliminate the 5 in r4c1, which left the only 5 in that column at r7c1. Once I got that 5 solved, everything else fell into place.
 
 But yeah, I like the (237) xy-wing much better, very elegant. I will get the hang of it one day!
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		| Marty R. 
 
 
 Joined: 12 Feb 2006
 Posts: 5770
 Location: Rochester, NY, USA
 
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				|  Posted: Fri Mar 06, 2009 4:57 am    Post subject: |   |  
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				| Hi Fungii, 
 Welcome to the forum. Enjoy your stay, lots of funky elegance here!!
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		| crunched 
 
 
 Joined: 05 Feb 2008
 Posts: 168
 
 
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				|  Posted: Fri Mar 06, 2009 6:57 am    Post subject: |   |  
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				| So how did your technique solve the puzzle? I did notice that the 7s in column 3 force an elimination of the rest of the 7s in box 4. Also, more eliminations were made in box 6.
 See resulting grid below.
 But I could not go any further.
 How did you take another step to get rid of the 5 in r4c1?
 
 
 
 
  	  | Fungii wrote: |  	  | I'm kind of new to these funky techniques you guys use to solve these puzzles, so I didn't see the (237) xy-wing.   
 Instead, I worked on the (5,7) r5c3. Both options worked to eliminate the 5 in r4c1, which left the only 5 in that column at r7c1. Once I got that 5 solved, everything else fell into place.
 
 But yeah, I like the (237) xy-wing much better, very elegant. I will get the hang of it one day!
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  	  | Code: |  	  | +-------------+-------------+-----------+
 | 9   247 1   | 3   6   27  | 5 28  48  |
 | 8   5   6   | 4   1   29  | 7 239 39  |
 | 27  247 3   | 5   8   279 | 6 1   49  |
 +-------------+-------------+-----------+
 | 235 239 257 | 279 259 4   | 8 6   1   |
 | 6   1   57  | 79  3   8   | 4 59  2   |
 | 4   29  8   | 6   259 1   | 3 7   59  |
 +-------------+-------------+-----------+
 | 125 8   25  | 12  7   3   | 9 4   6   |
 | 137 37  9   | 18  4   6   | 2 358 358 |
 | 23  6   4   | 289 29  5   | 1 38  7   |
 +-------------+-------------+-----------+
 
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 Play this puzzle online at the Daily Sudoku site
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		| Charlie W 
 
 
 Joined: 26 Feb 2009
 Posts: 1
 
 
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				|  Posted: Fri Mar 06, 2009 1:21 pm    Post subject: |   |  
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				| Crunched, if r5c3 is 5, then r4c1 is not 5. If r5c3 is 7, then r4c3 and r4c5 are both 25 (via r5c4). These are matched pairs that also eliminate 5 in r4c1. 
 I'm just getting the hang of these VHs (helped a lot by all the archive puzzles and posts - thanks everyone), and I also used the xy wing for this one.
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		| crunched 
 
 
 Joined: 05 Feb 2008
 Posts: 168
 
 
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				|  Posted: Fri Mar 06, 2009 2:07 pm    Post subject: |   |  
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				| Ahhh. Very good. I see the logic in what you did here to eliminate the 5 in r4c1. Thanks.
 
 
  	  | Charlie W wrote: |  	  | Crunched, if r5c3 is 5, then r4c1 is not 5. If r5c3 is 7, then r4c3 and r4c5 are both 25 (via r5c4). These are matched pairs that also eliminate 5 in r4c1. 
 I'm just getting the hang of these VHs (helped a lot by all the archive puzzles and posts - thanks everyone), and I also used the xy wing for this one.
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