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williamholiday
Joined: 01 Apr 2011
Posts: 28
Location: Connecticut USA
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 Posted: Tue Apr 05, 2016 5:30 am Post subject: Apr 5 VH |
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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 |
+-----------+--------+---------+
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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
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| Posted: Tue Apr 05, 2016 3:10 pm Post subject: |
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| 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
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| Posted: Tue Apr 05, 2016 5:22 pm Post subject: |
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| 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
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| Posted: Tue Apr 05, 2016 5:56 pm Post subject: |
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| Does anybody have a fourth! |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Tue Apr 05, 2016 7:49 pm Post subject: |
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| 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
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| Posted: Tue Apr 05, 2016 11:37 pm Post subject: |
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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
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| Posted: Wed Apr 06, 2016 1:04 am Post subject: |
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| 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
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| Posted: Wed Apr 06, 2016 1:39 am Post subject: |
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| 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
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Wed Apr 06, 2016 2:04 am Post subject: |
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| 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
Posts: 2
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| Posted: Wed Apr 06, 2016 8:19 am Post subject: |
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| 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
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| Posted: Wed Apr 06, 2016 11:36 am Post subject: |
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| 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
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| Posted: Wed Apr 06, 2016 11:04 pm Post subject: |
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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).  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
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| Posted: Wed Apr 06, 2016 11:37 pm Post subject: |
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| 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
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| Posted: Wed Apr 27, 2016 1:54 pm Post subject: BUG +1 |
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| 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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