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		Earl
 
 
  Joined: 30 May 2007 Posts: 677 Location: Victoria, KS
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				 Posted: Sat Nov 14, 2009 3:39 pm    Post subject: Nov 14 DB | 
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				The Nov 14 DB is a refreshing challenge.  Different paths present themselves.  Here is a one-stepper.  Norm will surely find another.
 
 
A Solution:  type 4 UR 25 where either solution makes R8C7 <2>.  The type I UR 89 does not solve the puzzle.
 
Earl
 
 
 
 	  | Code: | 	 		  
 
+-------+-------+-------+
 
| . . . | 1 . . | . 4 5 |
 
| . 3 . | . . . | 7 . 9 |
 
| . . . | . . 3 | . 8 2 |
 
+-------+-------+-------+
 
| 3 . 7 | 5 9 . | 4 . . |
 
| . . . | . 4 . | . . . |
 
| . . 5 | . 3 6 | 8 . 7 |
 
+-------+-------+-------+
 
| 7 4 . | 3 . . | . . . |
 
| 5 . 6 | . . . | . 3 . |
 
| 2 9 . | . . 5 | . . . |
 
+-------+-------+-------+
 
 | 	  
 
Play this puzzle online at the Daily Sudoku site | 
			 
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		arkietech
 
 
  Joined: 31 Jul 2008 Posts: 1834 Location: Northwest Arkansas USA
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				 Posted: Sat Nov 14, 2009 6:08 pm    Post subject:  | 
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				An xy-chain solution: 	  | Code: | 	 		   *-----------------------------------------------------------*
 
 | 689   7     89    | 1     268   29    | 3     4     5     |
 
 | 168   3     2     | 68    5     4     | 7     16    9     |
 
 | 169   5     4     |c69    7     3     |d16    8     2     |
 
 |-------------------+-------------------+-------------------|
 
 | 3     26    7     | 5     9     8     | 4     126  *16    |
 
 | 89    26    89    | 7     4     1     | 25    256   3     |
 
 | 4     1     5     | 2     3     6     | 8     9     7     |
 
 |-------------------+-------------------+-------------------|
 
 | 7     4     1     | 3     68    29    | 259   25    8-6   |
 
 | 5     8     6     |b49    12    7     | 29    3    a14    |
 
 | 2     9     3     | 468   168   5     |*16    7     148-6 |
 
 *-----------------------------------------------------------*
 
(6=1)r4c9-(1=4)r8c9-(4=9)r8c4-(9=6)r3c4-(6=1)r3c7-(1=6)r9c7 
 
=> r79c9<>6;
 
singles | 	 
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		nataraj
 
 
  Joined: 03 Aug 2007 Posts: 1048 Location: near Vienna, Austria
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				 Posted: Sun Nov 15, 2009 12:09 pm    Post subject:  | 
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				This one could very well be found in the "one trick pony" section (uniqueness) ...
 
 
My approch was slightly different from Earl's:
 
The UR 26 (r45c28) with strong link on 2 in row 4 means that r5c8 cannot be "6". 
 
This turns the UR 25 (r57c78) into a type 1 and solves r7c7 ("9") and the puzzle. | 
			 
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		tlanglet
 
 
  Joined: 17 Oct 2007 Posts: 2468 Location: Northern California Foothills
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				 Posted: Sun Nov 15, 2009 2:47 pm    Post subject:  | 
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				Hi Nataraj, good to see that you still are able to visit occasionally.
 
 
I also used the UR26 in r45c28. To prevent the deadly patter, either r4c8=1 or r5c8=5. This strong inference can be used as follows:
 
 
(6=1)r4c9 - UR26[(1)r4c8 = (5)r5c8] - (5=2)r5c7 - (2=9)r8c7 - (9=4)r8c4 - (4=1)r8c9 - (1=6)r9c7; r79c9<>6 to complete the puzzle.
 
 
(Dan, I just read your post and realize that we used two different chains to form identical pincers. Interesting..........)
 
 
Ted | 
			 
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		keith
 
 
  Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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				 Posted: Sun Nov 15, 2009 4:30 pm    Post subject:  | 
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				I had a type 6 UR that put 5's on a diagonal, opening up a useful type 1 UR.
 
 
Edit:  After basics:
 
 
 	  | Code: | 	 		  +----------------+----------------+----------------+
 
| 689  7    89   | 1    268  29   | 3    4    5    | 
 
| 168  3    2    | 68   5    4    | 7    16   9    | 
 
| 169  5    4    | 69   7    3    | 16   8    2    | 
 
+----------------+----------------+----------------+
 
| 3    26   7    | 5    9    8    | 4    126# 16   | 
 
| 89   26   89   | 7    4    1    | 25@  256  3    | 
 
| 4    1    5    | 2    3    6    | 8    9    7    | 
 
+----------------+----------------+----------------+
 
| 7    4    1    | 3    68   29   | 259  25@  68   | 
 
| 5    8    6    | 49   12   7    | 29   3    14   | 
 
| 2    9    3    | 468  168  5    | 16   7    1468 | 
 
+----------------+----------------+----------------+ | 	  
 
I did not notice the 89 UR in B14.  Rather, there is a Type 6 UR 25 in B69.  It solves the cells @ as 5.  This reveals a Type 1 UR 26 in B46 that solves # as 1.
 
 
Kind of cool.  It is rare to find a Type 6 UR that is of much help.
 
 
Keith | 
			 
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		keith
 
 
  Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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				 Posted: Wed Nov 18, 2009 3:43 pm    Post subject: Re: Nov 14 DB | 
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				 	  | Earl wrote: | 	 		  The Nov 14 DB is a refreshing challenge.  Different paths present themselves.  Here is a one-stepper.  Norm will surely find another.
 
 
A Solution:  type 4 UR 25 where either solution makes R8C7 <2>.  The type I UR 89 does not solve the puzzle.
 
Earl | 	  
 
Earl,
 
 
I don't see this UR.
 
 
Keith | 
			 
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