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roaa
Joined: 18 Apr 2009 Posts: 111 Location: Sweden

Posted: Wed Jul 02, 2014 8:03 am Post subject: JUL 2 VH 


The usual! The XYwing 467 forces r7c5 and r9c5 <>7 which is enough. 

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hughwill
Joined: 05 Apr 2010 Posts: 391 Location: Birmingham UK

Posted: Wed Jul 02, 2014 8:36 am Post subject: Jul 2 VH 


After basics:
Code: 
++++
 6 8 5  2 47 9  134 347 137 
 3 9 4  567 167 15  2 67 8 
 2 7 1  46 8 3  46 5 9 
++++
 8 5 3  1 2 4  7 9 6 
 4 1 7  3 9 6  58 28 25 
 9 2 6  8 5 7  134 34 13 
++++
 5 4 8  67 367 2  9 1 37 
 7 36 9  45 14 15  368 368 2 
 1 36 2  9 37 8  5 367 4 
++++

Play this puzzle online at the Daily Sudoku site
Same answer jumped out. Didn't even need to look for the 67 WWing
in c5 setting r2c4 to 5!
Hugh 

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Marty R.
Joined: 12 Feb 2006 Posts: 5609 Location: Rochester, NY, USA

Posted: Wed Jul 02, 2014 2:53 pm Post subject: Re: Jul 2 VH 


hughwill wrote:  After basics:
Code: 
++++
 6 8 5  2 47 9  134 347 137 
 3 9 4  567 167 15  2 67 8 
 2 7 1  46 8 3  46 5 9 
++++
 8 5 3  1 2 4  7 9 6 
 4 1 7  3 9 6  58 28 25 
 9 2 6  8 5 7  134 34 13 
++++
 5 4 8  67 367 2  9 1 37 
 7 36 9  45 14 15  368 368 2 
 1 36 2  9 37 8  5 367 4 
++++

Play this puzzle online at the Daily Sudoku site
Same answer jumped out. Didn't even need to look for the 67 WWing
in c5 setting r2c4 to 5!
Hugh 
Hugh,
I don't understand how the wing sets r2c4 to 5. The wing does some damage, but I don't see that conclusion. 

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hughwill
Joined: 05 Apr 2010 Posts: 391 Location: Birmingham UK

Posted: Wed Jul 02, 2014 5:56 pm Post subject: 


Hi Marty
I may be wrong here but:
If r2c8 = 6 then r2c5<>6, r7c5 must be 6 making r7c4=7
If r7c4=6 then r7c5<>6 and r2c5 must be 6 making r2c8=7
Either way the 67s act as a pair, so r2c4<>7 or 6 and so =5
Have I made an error?
Hugh 

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keith
Joined: 19 Sep 2005 Posts: 3287 Location: near Detroit, Michigan, USA

Posted: Wed Jul 02, 2014 8:07 pm Post subject: 


Hugh,
Your grid has unsolved basics in R5B6 and R8C8. And yes, the 67s are a remote pair, seen as a pair of wwings with a link on 6 in R3 and a link on 7 in C9.
Keith 

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Marty R.
Joined: 12 Feb 2006 Posts: 5609 Location: Rochester, NY, USA

Posted: Wed Jul 02, 2014 8:37 pm Post subject: 


hughwill wrote:  Hi Marty
I may be wrong here but:
If r2c8 = 6 then r2c5<>6, r7c5 must be 6 making r7c4=7
If r7c4=6 then r7c5<>6 and r2c5 must be 6 making r2c8=7
Either way the 67s act as a pair, so r2c4<>7 or 6 and so =5
Have I made an error?
Hugh 
I don't know if you've made an error or not because a glaring weakness of mine is the inability to follow along with that style of reasoning. However, your premise of either r2c8 or r7c4 being 6 puzzles me because I don't see anything that says either of those two cells must be 6. Perhaps someone else will come along and straighten me out.
But my original response was about the statement about the WWing, "Didn't even need to look for the 67 WWing
in c5 setting r2c4 to 5!"
The WWing consists of 67 cells in r2c8 and r7c4, connected by the strong link on 6 in c5. That says at least one of r2c8 and r7c4 must be 7, thus making r2c4<>7. 

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keith
Joined: 19 Sep 2005 Posts: 3287 Location: near Detroit, Michigan, USA

Posted: Wed Jul 02, 2014 9:18 pm Post subject: 


Quote:  connected by the strong link on 6 in c5  

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Lee
Joined: 10 Jun 2014 Posts: 24 Location: San Francisco

Posted: Wed Jul 02, 2014 10:25 pm Post subject: 


Hello Everyone,
Have to get in my two cents.
Looks to me like r2c4 is connected by two wwings both using the same end points (r7c4 and r2c8):
The first wwing involves a strong link on 6 in c5 as pointed out by Marty.
The second wwing involves a strong link on 7 in block 9 (r7c9 to r9c8).
The first wwing eliminates the 7 from r2c4. The second wwing eliminates the 6 from r2c4. Thus between the two we are left with a 5 in r2c4.
This seems pretty strange, but it also seems to be true, or am I really confused?
Comments??? 

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Marty R.
Joined: 12 Feb 2006 Posts: 5609 Location: Rochester, NY, USA

Posted: Wed Jul 02, 2014 11:23 pm Post subject: 


keith wrote:  Quote:  connected by the strong link on 6 in c5  
Keith,
What's the question? I may have missed the fact that there was a double WWing, but each 6 in c5 sees a 67 cell based on the posted grid. 

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keith
Joined: 19 Sep 2005 Posts: 3287 Location: near Detroit, Michigan, USA

Posted: Wed Jul 02, 2014 11:26 pm Post subject: 


Marty R. wrote:  keith wrote:  Quote:  connected by the strong link on 6 in c5  
Keith,
What's the question? I may have missed the fact that there was a double WWing, but each 6 in c5 sees a 67 cell based on the posted grid. 
Marty,
Thank you, I would not have recognized it as a strong link.
Keith 

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Marty R.
Joined: 12 Feb 2006 Posts: 5609 Location: Rochester, NY, USA

Posted: Wed Jul 02, 2014 11:30 pm Post subject: 


Quote:  This seems pretty strange, but it also seems to be true, or am I really confused? 
Sounds good to me. What really seems strange is that there are two strong links on both 6 and 7. 

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keith
Joined: 19 Sep 2005 Posts: 3287 Location: near Detroit, Michigan, USA

Posted: Thu Jul 03, 2014 1:33 am Post subject: 


Marty R. wrote:  Quote:  This seems pretty strange, but it also seems to be true, or am I really confused? 
Sounds good to me. What really seems strange is that there are two strong links on both 6 and 7. 
Marty,
The double Wwing is not that rare. What's more, it always makes a cycle, in which all links become strong.
Code:  ++++
 6 8 5  2 47 9 134 347 e137
 3 9 4 567 167 15  2 d67 8 
 2 7 1 b46 8 3 c46 5 9 
++++
 8 5 3  1 2 4  7 9 6 
 4 1 7  3 9 6  8 2 5 
 9 2 6  8 5 7  134 34 13 
++++
 5 4 8 a67 367 2  9 1 f37 
 7 36 9  45 14 15  36 8 2 
 1 36 2  9 37 8  5 367 4 
++++

In this case a6b6c6d7e7f7a, in my pidgin notation
In other words, buddies of a and b cannot be 6, of f and a cannot be 7, etc. I have marked the eliminations I can see.
Some Mwings are also often cycles.
Here is another XY cycle in the same grid, found by Sudoku Susser:
Code:  Found a 7link Simple Forcing Loop. If we assume that square [x=2,y=9] is <3> then we can make the following chain of conclusions:
[x=5,y=9] must be <7>, which means that
[x=4,y=7] must be <6>, which means that
[x=4,y=3] must be <4>, which means that
[x=7,y=3] must be <6>, which means that
[x=7,y=8] must be <3>, which means that
[x=2,y=8] must be <6>, which means that
[x=2,y=9] must be <3>.
This forms a inherently bidirectional loop through the puzzle that permits reductions. On each edge of the loop, one of the two squares must have a particular value, so their common buddies cannot contain that value, as follows:
One of [x=2,y=9] and [x=5,y=9] must be <3>.
One of [x=5,y=9] and [x=4,y=7] must be <7>.
One of [x=4,y=7] and [x=4,y=3] must be <6>.
One of [x=4,y=3] and [x=7,y=3] must be <4>.
One of [x=7,y=3] and [x=7,y=8] must be <6>.
One of [x=7,y=8] and [x=2,y=8] must be <3>.
One of [x=2,y=8] and [x=2,y=9] must be <6>.
Thus we can deduce that:
[x=8,y=9]  cannot contain <3> because of [x=2,y=9] and [x=5,y=9].
[x=5,y=7]  cannot contain <7> because of [x=5,y=9] and [x=4,y=7].
[x=4,y=2]  cannot contain <6> because of [x=4,y=7] and [x=4,y=3].

Keith 

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hughwill
Joined: 05 Apr 2010 Posts: 391 Location: Birmingham UK

Posted: Thu Jul 03, 2014 7:10 pm Post subject: 


Marty wrote:
Quote:  I don't know if you've made an error or not because a glaring
weakness of mine is the inability to follow along with that style of reasoning.
However, your premise of either r2c8 or r7c4 being 6 puzzles me because I
don't see anything that says either of those two cells must be 6. Perhaps
someone else will come along and straighten me out. 
Yes Marty you've seen the flaw in my logic of course (without the
strong link in C9 on 7 from later contributors) there is no reason
why one of these squares must be 6. I might have got the right
answer, but through wrong reasoning!
Hugh 

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keith
Joined: 19 Sep 2005 Posts: 3287 Location: near Detroit, Michigan, USA

Posted: Fri Jul 04, 2014 1:33 pm Post subject: 


Note the Mwing that solves it in one step: Code:  ++++
 6 8 5  2 47 9  134 347 137 
 3 9 4 B567 167 15  2 A67 8 
 2 7 1  46 8 3  46 5 9 
++++
 8 5 3  1 2 4  7 9 6 
 4 1 7  3 9 6  8 2 5 
 9 2 6  8 5 7  134 34 13 
++++
 5 4 8 C67 D367 2  9 1 37 
 7 36 9  45 14 15  36 8 2 
 1 36 2  9 37 8  5 367 4 
++++  The logic is:
If A is 7, B is not 7, C is 7. If C is not 6, D is 6. AD are pincers on 6, taking 6 from R2C5 and solving the puzzle..
Keith 

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Marty R.
Joined: 12 Feb 2006 Posts: 5609 Location: Rochester, NY, USA

Posted: Fri Jul 04, 2014 3:11 pm Post subject: 


Now that's something that I do see frequentlyRemote Pairs being able to be played as both M and WWings. 

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