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		daj95376
 
 
  Joined: 23 Aug 2008 Posts: 3854
 
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				 Posted: Wed Jun 08, 2011 3:46 pm    Post subject: Puzzle 11/06/08: ~ XY (BBDB?) | 
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				 	  | Code: | 	 		   +-----------------------+
 
 | . 9 . | . 7 . | 2 8 . |
 
 | 8 . . | . . 2 | . . . |
 
 | . . 4 | . . 8 | 3 7 . |
 
 |-------+-------+-------|
 
 | . . . | 7 . . | 1 . . |
 
 | 2 . . | . 8 1 | . . 4 |
 
 | . 1 9 | . 4 6 | . . . |
 
 |-------+-------+-------|
 
 | 9 . 1 | 6 . . | . 4 8 |
 
 | 6 . 2 | . . . | 9 . . |
 
 | . . . | . 9 . | 6 . 3 |
 
 +-----------------------+
 
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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: Thu Jun 09, 2011 8:01 am    Post subject:  | 
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				Two steps - the second a neat als move (I think)
 
  	  | Quote: | 	 		  w-wing(51) ; (5=1)r3c1 - r3c5=r2c5 - (1=5)r2c8 ; r2c23<>5, r3c9<>5 
 
hp(15)r13c1=(5)r1c3 - (5=3)r1c6 ; r1c1<>3   
 
... some might prefer the naked triple(367) ... | 	 
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		tlanglet
 
 
  Joined: 17 Oct 2007 Posts: 2468 Location: Northern California Foothills
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				 Posted: Thu Jun 09, 2011 2:29 pm    Post subject:  | 
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				An anp() that makes two distinct deletions by continuing the chain; both deletions are required to complete the puzzle in one step.
 
 
anp(15=3)r13c1-(3=5)r1c6*-r789c6=r8c5-r8c89=r7c7-r5c7=(5)r6c9; r1c3*, r6c1<>5
 
 
Ted | 
			 
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		tlanglet
 
 
  Joined: 17 Oct 2007 Posts: 2468 Location: Northern California Foothills
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				 Posted: Thu Jun 09, 2011 2:38 pm    Post subject:  | 
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				 	  | peterj wrote: | 	 		  Two steps - the second a neat als move (I think)
 
  	  | Quote: | 	 		  w-wing(51) ; (5=1)r3c1 - r3c5=r2c5 - (1=5)r2c8 ; r2c23<>5, r3c9<>5 
 
hp(15)r13c1=(5)r1c3 - (5=3)r1c6 ; r1c1<>3   
 
... some might prefer the naked triple(367) ... | 	 
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Peter, Interesting that we both viewed the same pattern thru different glasses. I was specifically looking for anp()s, and being methodical, I started looking in box 1 where I hit the jackpot.
 
 
And yes, the als was "neat"...........
 
 
Ted | 
			 
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		dejsmith
 
 
  Joined: 23 Oct 2005 Posts: 42
 
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				 Posted: Thu Jun 09, 2011 8:22 pm    Post subject:  | 
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				Where's Marty?  Documenting these alternative solutions stresses me out!  Don't worry, though, I'll fall a couple of days behind soon.
 
 
1. Coloring on 7: r8c2<>7 (not needed, just 1st thing I saw)
 
2. Same W Wing as Peter with same eliminations
 
3. X Wing on 3 r25; r14c3<>3
 
4. XYZ Wing 135; r1c3<>5
 
 
Dave | 
			 
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		daj95376
 
 
  Joined: 23 Aug 2008 Posts: 3854
 
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				 Posted: Thu Jun 09, 2011 11:12 pm    Post subject:  | 
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				Hmmm!!! I guess that I'm the only one who doesn't see Peter's ALS move.
 
 
 	  | Code: | 	 		   +--------------------------------------------------------------+
 
 |  15+3  9     356   |  4     7     35    |  2     8     156   |
 
 |  8     67    367   |  359   16    2     |  4     15    1569  |
 
 |  15    2     4     |  59    16    8     |  3     7     169   |
 
 |--------------------+--------------------+--------------------|
 
 |  345   458   358   |  7     35    9     |  1     6     2     |
 
 |  2     567   3567  |  35    8     1     |  57    9     4     |
 
 |  57    1     9     |  2     4     6     |  8     3     57    |
 
 |--------------------+--------------------+--------------------|
 
 |  9     357   1     |  6     2     357   |  57    4     8     |
 
 |  6     3457  2     |  8     35    3457  |  9     15    157   |
 
 |  457   4578  578   |  1     9     457   |  6     2     3     |
 
 +--------------------------------------------------------------+
 
 # 62 eliminations remain
 
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I see: (15)r13c1=(3)r1c1
 
 
I don't see: (15)r13c1=(5)r1c3 because r1c1=3 and r3c1=5 force the left side false but not the right side true.
 
 
 
Ted: Nice chain, but it has a shorter cousin.
 
 
anp(15=3)r13c1-(3=5)r1c6*-r789c6=r8c5-r4c5=(5)r4c123; r1c3*, r6c1<>5
 
 
 
Upon reviewing my solver's solution, I had no idea that the XYZ-Wing and W-Wing were all that's necessary to crack this puzzle. | 
			 
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		Luke451
 
 
  Joined: 20 Apr 2008 Posts: 310 Location: Southern Northern California
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				 Posted: Fri Jun 10, 2011 12:37 am    Post subject:  | 
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				 	  | daj95376 wrote: | 	 		  | Hmmm!!! I guess that I'm the only one who doesn't see Peter's ALS move. | 	  
 
 
Peter called it an ALS but notated it as a hidden pair. With the 1s conjugate it seems hp(15)r13c1=(5)r1c3 is a valid statement. | 
			 
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		ronk
 
 
  Joined: 07 May 2006 Posts: 398
 
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				 Posted: Fri Jun 10, 2011 12:58 am    Post subject:  | 
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				 	  | daj95376 wrote: | 	 		  | Hmmm!!! I guess that I'm the only one who doesn't see Peter's ALS move. | 	  
 
Did you let the "ALS" mislead you? It's actually an AHS. | 
			 
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		daj95376
 
 
  Joined: 23 Aug 2008 Posts: 3854
 
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				 Posted: Fri Jun 10, 2011 5:04 am    Post subject:  | 
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				I prefer to have the conjugate on <1> spelled out.
 
 
(15=13)r31c1 - (3=5)r1c5 - (5)r1c13 = (5)r3c1 - (13=15)r13c1 => r1c3,r469c1<>5 , r1c1<>3 | 
			 
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		peterj
 
 
  Joined: 26 Mar 2010 Posts: 974 Location: London, UK
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				 Posted: Fri Jun 10, 2011 7:12 am    Post subject:  | 
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				Apologies for the confusion.   
 
Though I almost wrote it for effect as...       
 
 	  | Code: | 	 		  | hp(15)r13c1=np(35)r1c36 | 	 
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		ronk
 
 
  Joined: 07 May 2006 Posts: 398
 
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				 Posted: Fri Jun 10, 2011 12:09 pm    Post subject:  | 
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				 	  | daj95376 wrote: | 	 		  I prefer to have the conjugate on <1> spelled out.
 
 
(15=13)r31c1 - (3=5)r1c5 - (5)r1c13 = (5)r3c1 - (13=15)r13c1 => r1c3,r469c1<>5 , r1c1<>3 | 	  
 
Notation of hidden sets has always been a bit awkward IMO, but cells appearing three times is a new record.
 
 
Unless the size of the AHS is much smaller than the complementary ALS, I prefer using the ALS: (3=5)r1c6 - (5=673)r1c3,r2c23 ==> r1c1<>3 | 
			 
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		daj95376
 
 
  Joined: 23 Aug 2008 Posts: 3854
 
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				 Posted: Fri Jun 10, 2011 3:55 pm    Post subject:  | 
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				Peter, no need for apologies. As Ron just said, "Notation of hidden sets has always been a bit awkward IMO". Since I've been bitten before on the notation, I was sensitive about being caught off balance again.
 
 
I agree that my solution sucks! _   _
 
 
Ron, how about: (it uses the conjugate on <1>)
 
 
(3=5)r1c5 - (5)r1c13 = (5-1)r3c1 = (1)r1c1 => r1c1<>3
 
 
This gets around reading Peter's chain from r-to-l and performing -(5)r1c3 instead of -(5)r1c13.
 
 
Regards, Danny | 
			 
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		Luke451
 
 
  Joined: 20 Apr 2008 Posts: 310 Location: Southern Northern California
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				 Posted: Fri Jun 10, 2011 4:18 pm    Post subject:  | 
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				 	  | ronk wrote: | 	 		  | Unless the size of the AHS is much smaller than the complementary ALS, I prefer using the ALS: (3=5)r1c6 - (5=673)r1c3,r2c23 ==> r1c1<>3 | 	   
 
Seems that all along we've been looking at an ALS-xz   .
 
 
 	  | Code: | 	 		   +--------------------------------------------------------------+
 
 |  153   9     356   |  4     7     35    |  2     8     156   |
 
 |  8     67    367   |  359   16    2     |  4     15    1569  |
 
 |  15    2     4     |  59    16    8     |  3     7     169   |
 
 |--------------------+--------------------+--------------------|
 
 |  345   458   358   |  7     35    9     |  1     6     2     |
 
 |  2     567   3567  |  35    8     1     |  57    9     4     |
 
 |  57    1     9     |  2     4     6     |  8     3     57    |
 
 |--------------------+--------------------+--------------------|
 
 |  9     357   1     |  6     2     357   |  57    4     8     |
 
 |  6     3457  2     |  8     35    3457  |  9     15    157   |
 
 |  457   4578  578   |  1     9     457   |  6     2     3     |
 
 +--------------------------------------------------------------+  | 	 
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		Marty R.
 
 
  Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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				 Posted: Sat Jun 11, 2011 4:20 am    Post subject:  | 
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				 	  | Quote: | 	 		  | Upon reviewing my solver's solution, I had no idea that the XYZ-Wing and W-Wing were all that's necessary to crack this puzzle. | 	  
 
That's what I did, the W-Wing followed by the XYZ-Wing. | 
			 
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