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		daj95376
 
 
  Joined: 23 Aug 2008 Posts: 3854
 
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				 Posted: Tue May 31, 2011 8:37 pm    Post subject: Puzzle 11/05/31: ~ XY | 
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				 	  | Code: | 	 		   +-----------------------+
 
 | 6 8 . | 5 3 . | . . . |
 
 | 5 . . | . . 2 | . . . |
 
 | . . 3 | 8 7 . | 1 . . |
 
 |-------+-------+-------|
 
 | 7 . 4 | . . 8 | 6 9 . |
 
 | 1 . 6 | . 5 9 | . . . |
 
 | . 9 . | 7 6 3 | . . . |
 
 |-------+-------+-------|
 
 | . . 5 | 3 . . | . . 9 |
 
 | . . . | 2 . . | . 8 . |
 
 | . . . | . . . | 5 . 7 |
 
 +-----------------------+
 
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Play this puzzle online at the Daily Sudoku site | 
			 
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		peterj
 
 
  Joined: 26 Mar 2010 Posts: 974 Location: London, UK
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				 Posted: Wed Jun 01, 2011 9:24 am    Post subject:  | 
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				<ott>
 
For fun ... notice there is an almost xy-wing(79-1) pivot at r8c3 and this eliminates (1)r2c5 which makes a w-wing(16) in band 3 whole.. conveniently a bivalue (1=6) makes the xywing and the same eliminations as the wwing..
 
 	  | Code: | 	 		  | (6=1)r8c9 - (1)r8c3=xywing(79-1)r8c3,r2c3,r8c3 - (1)r2c5=wwing(16)r7c2,r8c9 ; r8c2<>6, r7c8<>6 | 	  
 
This would be considered inflammatory on "another board"! Hopefully here it will pass off as intersting/amusing and of course OTT for this puzzle!
 
</ott> | 
			 
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		tlanglet
 
 
  Joined: 17 Oct 2007 Posts: 2468 Location: Northern California Foothills
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				 Posted: Wed Jun 01, 2011 2:10 pm    Post subject:  | 
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				Peter, very    find!
 
 
Here is an interesting twist using an identical anp() but reversing the order to solve the puzzle.  
 
 
anp(12=9)r9c23-als(9=2)r821c3-(2-4)r3c2-(4=6)r3c6-(6=5)r3c8-(5=1)r6c8; r9c8<>1
 
 
anp(9=12)r9c32-(1=6)r7c2-r7c8=r3c8-r3c6=(6-1)r9c6; r9c6<>1
 
 
Ted | 
			 
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		ronk
 
 
  Joined: 07 May 2006 Posts: 398
 
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				 Posted: Wed Jun 01, 2011 2:29 pm    Post subject:  | 
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				 	  | peterj wrote: | 	 		  For fun ... notice there is an almost xy-wing(79-1) pivot at r8c3 and this eliminates (1)r2c5 which makes a w-wing(16) in band 3 whole.. conveniently a bivalue (1=6) makes the xywing and the same eliminations as the wwing..
 
 	  | Code: | 	 		  | (6=1)r8c9 - (1)r8c3=xywing(79-1)r8c3,r2c3,r8c3 - (1)r2c5=wwing(16)r7c2,r8c9 ; r8c2<>6, r7c8<>6 | 	 
  | 	  
 
Haven't figured out how one goes about finding such a thing, but it definitely is interesting. It can be viewed as a chain which doubly-links [an AALS.]
 
 
(6=1)r8c9 - (1=79)als:r8c35 - (79=14)aals:r2c35 - (4=1)r7c5 - (1=6)r7c2 ==> r7c8, r8c2<>6
 
 
When viewed as a w-wing, there is a derived strong inference (1) r8c35 = (1)r7c5 linking the two bivalues.
 
 
[edit: Not sure how one would read the aals chain from right-to-left though.]
  Last edited by ronk on Wed Jun 01, 2011 5:08 pm; edited 1 time in total | 
			 
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		daj95376
 
 
  Joined: 23 Aug 2008 Posts: 3854
 
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				 Posted: Wed Jun 01, 2011 4:33 pm    Post subject:  | 
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				 	  | Code: | 	 		  after basics
 
 +--------------------------------------------------------------+
 
 |  6     8     127   |  5     3     14    |  9     27    24    |
 
 |  5     147   17    |  69    149   2     |  78    3     468   |
 
 |  9     24    3     |  8     7     46    |  1     56    2456  |
 
 |--------------------+--------------------+--------------------|
 
 |  7     5     4     |  1     2     8     |  6     9     3     |
 
 |  1     3     6     |  4     5     9     |  78    27    28    |
 
 |  2     9     8     |  7     6     3     |  4     15    15    |
 
 |--------------------+--------------------+--------------------|
 
 |  8     16    5     |  3     14    7     |  2     146   9     |
 
 |  4     167   179   |  2     19    5     |  3     8     16    |
 
 |  3     12    129   |  69    8     146   |  5     14    7     |
 
 +--------------------------------------------------------------+
 
 # 42 eliminations remain
 
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My perspective on peterj's solution:
 
 
If r8c3=1, then (1=6)r7c2,r8c9 ... and the eliminations follow.
 
 
Else r8c3<>1, then two steps: XY-Wing, W-Wing ... and the eliminations follow.
 
 
 
 
 
My solver encountered similar logic: just less interesting
 
 
If r2c9=4, then r2c9<>6.
 
 
Else r2c9<>4 and this chain exists:
 
 
(6)r8c9 = (6-7)r8c2 = (7-4)r2c2 = r2c5 - (4=1)r7c5 - (1=6)r7c2 - r8c2 = (6)r8c9  =>  r2c9<>6 | 
			 
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		peterj
 
 
  Joined: 26 Mar 2010 Posts: 974 Location: London, UK
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				 Posted: Wed Jun 01, 2011 5:35 pm    Post subject:  | 
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				 	  | ronk wrote: | 	 		  | Haven't figured out how one goes about finding such a thing, but it definitely is interesting.  | 	  
 
fwiw I have spent the last year on this forum learning patterns, I now find I see almost-patterns relatively naturally (which some might think worrying!). In this case I saw the aw-wing and noticed the 1 was an elimination of the axy-wing - it just read better the other way round! 
 
 
I know they are nets really but I dont find them that way.
 
 
farpointer/kobold on the au forum used them a lot - especially fish - find those harder.
 
 
I haven't figured how people find really long AIC with multiple digits and hidden sets - other than just legthy "fishing" I guess. | 
			 
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		Marty R.
 
 
  Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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				 Posted: Thu Jun 02, 2011 4:17 am    Post subject:  | 
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				Two M-Wings
 
 
12; r9c6<>1
 
16, flightless with transport; r6c8, r8c9<>1 | 
			 
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