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Apr 5 VH

 
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williamholiday



Joined: 01 Apr 2011
Posts: 28
Location: Connecticut USA

PostPosted: Tue Apr 05, 2016 5:30 am    Post subject: Apr 5 VH Reply with quote

After basics:
Code:

+-----------+--------+---------+
| 9   7  6  | 2  8 3 | 5  4 1  |
| 8   14 14 | 5  7 6 | 9  2 3  |
| 5   2  3  | 1  4 9 | 8  7 6  |
+-----------+--------+---------+
| 4   8  2  | 6  3 5 | 1  9 7  |
| 6   9  5  | 8  1 7 | 24 3 24 |
| 37  13 17 | 4  9 2 | 6  5 8  |
+-----------+--------+---------+
| 23  5  49 | 39 6 8 | 7  1 24 |
| 237 34 8  | 37 5 1 | 24 6 9  |
| 1   6  79 | 79 2 4 | 3  8 5  |
+-----------+--------+---------+

Play this puzzle online at the Daily Sudoku site

xy wing 34-2 pivot in r8c2, r9c7<>2
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Tue Apr 05, 2016 3:10 pm    Post subject: Reply with quote

Interesting. There's another one pivoted in the same box, a 49-3 in r7c3.
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dongrave



Joined: 06 Mar 2014
Posts: 568

PostPosted: Tue Apr 05, 2016 5:22 pm    Post subject: Reply with quote

And there's also the 23-4 XY Wing also in the same box!
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williamholiday



Joined: 01 Apr 2011
Posts: 28
Location: Connecticut USA

PostPosted: Tue Apr 05, 2016 5:56 pm    Post subject: Reply with quote

Does anybody have a fourth!
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Tue Apr 05, 2016 7:49 pm    Post subject: Reply with quote

Quote:
Does anybody have a fourth!


Not a 4th XY, but there's an XYZ-Wing (237) hinged in r8c1=> -3r8c2
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Don W.



Joined: 05 Apr 2016
Posts: 2

PostPosted: Tue Apr 05, 2016 11:37 pm    Post subject: Reply with quote

There's also a W-Wing (possibly related to the XYZ-Wing): r6c1 and r8c4 cannot both be 3 because of column 2, so -7r8c1.

For yet another advanced solving approach, here's one I use sometimes but haven't seen a name for: if r7c9=2 and r8c7=4, then r6c1 and r7c3 both =3.

(At first I thought this was what people meant by XY-Wing, but it's different since it uses a pair of linked cells as the pivot. The general rule is: if two linked cells are limited to XY, and one of them is linked to a cell XZ and the other is linked to a cell YZ, and the XZ/YZ cells are linked, then the original pair of XYs can be resolved only one way.)
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Wed Apr 06, 2016 1:04 am    Post subject: Reply with quote

Quote:
For yet another advanced solving approach, here's one I use sometimes but haven't seen a name for: if r7c9=2 and r8c7=4, then r6c1 and r7c3 both =3.


Do, you've described an XY-Chain, but I don't see where a conclusion can be drawn. Yes, it's established that if r7c9=2, those other two cells are =3. But what if r7c9<>3? Usually to draw a conclusion there has to be a common outcome from the two premises.
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dongrave



Joined: 06 Mar 2014
Posts: 568

PostPosted: Wed Apr 06, 2016 1:39 am    Post subject: Reply with quote

Hi Marty, Don W's point was that if r7c9=2 then you can arrive at both r7c1=3 and r8c2=3 which cannot be the case since they're both in box 7 therefore r7c9 isn't 2 but of course this is just one of many ways to represent the XY-Wing. You could also represent it as the following (which I call the 'Van Hay' representation): r7c1 is 2 or it's 3; if it's 2 then r7c9=4 and if it's 3, then r8c2=4 so r8c7=2 so r7c9=4; therefore r7c9=4. Or you could represent it as a chain like if r7c9=2 then r7c1=3 so r8c2=4 so r8c7=2 so r7c9<>2 contradition. Or you could shift the chain's starting position to r7c1 and represent it as a chain with pincers on both ends as follows: (2=3)r7c1-(3=4)r8c2-(4=2)r8c7 => r7c9<>2. They're all just different ways of representing the same thing.
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Marty R.



Joined: 12 Feb 2006
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Location: Rochester, NY, USA

PostPosted: Wed Apr 06, 2016 2:04 am    Post subject: Reply with quote

Maybe I wasn't paying enough attention. I was looking at both r6c1 and r8c2 as =3.
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Don W.



Joined: 05 Apr 2016
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PostPosted: Wed Apr 06, 2016 8:19 am    Post subject: Reply with quote

Ah, I see. I'm effectively following three different connections in the cycle of four.
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Wed Apr 06, 2016 11:36 am    Post subject: Reply with quote

Code:
+-----------+--------+---------+
| 9   7  6  | 2  8 3 | 5  4 1  |
| 8   14 14 | 5  7 6 | 9  2 3  |
| 5   2  3  | 1  4 9 | 8  7 6  |
+-----------+--------+---------+
| 4   8  2  | 6  3 5 | 1  9 7  |
| 6   9  5  | 8  1 7 | 24 3 24 |
| 37  13 17 | 4  9 2 | 6  5 8  |
+-----------+--------+---------+
| 23  5  49 | 39 6 8 | 7  1 24 |
| 237 34 8  | 37 5 1 | 24 6 9  |
| 1   6  79 | 79 2 4 | 3  8 5  |
+-----------+--------+---------+


If we're gonna take the approach of solving based on an invalidity, let's take the short route.

(4=2*)r7c9-(2=4)r8c7-(4=3)r8c3-(3=2*)r7c1 contradiction=> -2r7c9
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dongrave



Joined: 06 Mar 2014
Posts: 568

PostPosted: Wed Apr 06, 2016 11:04 pm    Post subject: Reply with quote

Hey, wait a minute. That's not shorter; that's the same chain clockwise instead of anticlockwise (as my British mother-in-law would say). Very Happy Stop trying to trick me Marty. My favorites are still the chains with pincers on both ends that I learned about from you and Bat and Clement last year! They're the best!
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Wed Apr 06, 2016 11:37 pm    Post subject: Reply with quote

I agree. The real experts don't value solutions from contradictions as much as the more positive ones.
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George Woods



Joined: 28 Mar 2006
Posts: 304
Location: Dorset UK

PostPosted: Wed Apr 27, 2016 1:54 pm    Post subject: BUG +1 Reply with quote

Ok I have been away sometime so am doing some old ones - Given all the other solutions no-one has mentioned the BUG +1 which puts a 3 in r8c1
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