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Extending the vertex of a potential xy-wing

 
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tlanglet



Joined: 17 Oct 2007
Posts: 2461
Location: Northern California Foothills

PostPosted: Sun Dec 13, 2009 5:34 pm    Post subject: Extending the vertex of a potential xy-wing Reply with quote

Recently I stumbled onto the technique of extending the vertex of a potential xy-wing. Nothing is new about the approach, but I find it easy to perform and surprisingly fruitful.

Code:

 *--------------------------------------------------*
 | 4    12   23   | 8    5    9    | 136  7   f16   |
 |c39   6    7    | 1    2    4    |d39   8    5    |
 | 5    18   89   | 7    6    3    |e19   2    4    |
 |----------------+----------------+----------------|
 | 6    25   1    | 9    7    25   | 4    3    8    |
 | 7    9    4    | 3    8    1    | 5    6    2    |
 | 23   258  238  | 25   4    6    | 7    1    9    |
 |----------------+----------------+----------------|
 |b29   7    259  | 6    1    258  | 28   4    3    |
 | 8    3   a26   | 4    9    7    | 126  5    16   |
 | 1    4    256  | 25   3    258  | 268  9    7    |
 *--------------------------------------------------*


In box 7 noticed the two bivalue pairs 26 in r8c3, marked a, and 29 in r7c1, marked b. This is a potential xy-wing, but the needed bivalue pair 69 is not directly visible to either of the other bivalue pairs.

But what happens if we try to extend the vertex of the potential xy-wing? So, assume that the vertex is the 29 in r7c1 marked b. Thus, if r7c1=2 then r8c3=6 which will act as a pincer.

So the challenge is what happens if r7c1=9. Let's give it a try.
If r7c1=9 then r2c1=3. But if r2c1=3, then r2c7=9, which forces r3c7=1, which them makes r1c9=6.

So what to we have.
If r7c1=2 then r8c3=6, and
If r7c1=9 then r1c9=6 via the chain formed by extending the vertex of the potential xy=wing along the path marked bcdef.

Thus, we have two pincers for digit 6, which delete the 6 in r8c9. This pattern is best described as an AIC marked abcdef.
(6=2)r8c3 - (2=9)r7c1 - (9=3)r2c1 - (3=9)r2c7 - (9=1)r3c7 - (1=6)r1c9; r8c9<>6.

Again, I find this approach of extending the vertex of a potential xy-wing easy to perform and surprisingly fruitful.

Ted
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Marty R.



Joined: 12 Feb 2006
Posts: 5179
Location: Rochester, NY, USA

PostPosted: Sun Dec 13, 2009 7:39 pm    Post subject: Reply with quote

Ted, what my half-brain sees is a simple XY-Chain as the end result. It apparently started as a result of you seeing 2/3 of a potential XY-Wing, as opposed to a more random method of starting such a chain.

I've seen you a number of times make reference to a vertex extension. Have you made such extensions that result in pincers but don't have that XY-Chain pattern?
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keith



Joined: 19 Sep 2005
Posts: 3184
Location: near Detroit, Michigan, USA

PostPosted: Sun Dec 13, 2009 10:00 pm    Post subject: Reply with quote

Or, the -239 XY-wing solves the puzzle! Twisted Evil

Ted,

I see what you're doing, but I do not see a recipe. Is it this:

If you have two cells XY - YZ where X might be a pincer, start coloring on Z from YZ and see where that takes you?

Best wishes,

Keith
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tlanglet



Joined: 17 Oct 2007
Posts: 2461
Location: Northern California Foothills

PostPosted: Mon Dec 14, 2009 1:04 pm    Post subject: Reply with quote

Marty R. wrote:
Ted, what my half-brain sees is a simple XY-Chain as the end result. It apparently started as a result of you seeing 2/3 of a potential XY-Wing, as opposed to a more random method of starting such a chain.

I've seen you a number of times make reference to a vertex extension. Have you made such extensions that result in pincers but don't have that XY-Chain pattern?


Absolutely. The example that I included is an xy-chain, but other inferences do/can occur resulting in a true AIC.

As I mentioned in my post, nothing is new about this approach, but I find it easy to perform and surprisingly fruitful. I am not good at finding general chains so this technique is helpful to me.

Ted
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tlanglet



Joined: 17 Oct 2007
Posts: 2461
Location: Northern California Foothills

PostPosted: Mon Dec 14, 2009 1:07 pm    Post subject: Reply with quote

keith wrote:


Ted,

I see what you're doing, but I do not see a recipe. Is it this:

If you have two cells XY - YZ where X might be a pincer, start coloring on Z from YZ and see where that takes you?

Best wishes,

Keith


Sorry that the recipe was not describe better. Your definition is "right-on".

Ted
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daj95376



Joined: 23 Aug 2008
Posts: 3855

PostPosted: Mon Dec 14, 2009 4:32 pm    Post subject: Reply with quote

This is 10% academic and 90% rambling. Nothing I say is meant to imply that it's any better than another approach.

For me, certain techniques are all about the endpoint cells. If the endpoint cells don't generate an elimination, especially a productive one, then it's senseless to use any technique to connect them.

Consider the following:

Code:
 bivalue endpoint cells XZ & YZ ...       bivalue endpoint cells XZ & YZ ...   
 and their peer cells (p)                 and their peer cells (p)             
 +-----------------------------------+    +-----------------------------------+
 |  .  .  .  |  .  .  .  |  .  .  .  |    |  .  .  .  |  .  .  .  |  p  .  .  |
 |  p  .  .  |  .  .  .  |  .  . XZ  |    |  .  .  .  |  .  .  .  |  p  . XZ  |
 |  .  .  .  |  .  .  .  |  .  .  .  |    |  .  .  .  |  .  .  .  |  p  .  .  |
 |-----------+-----------+-----------|    |-----------+-----------+-----------|
 |  .  .  .  |  .  .  .  |  .  .  .  |    |  .  .  .  |  .  .  .  |  .  .  .  |
 |  .  .  .  |  .  .  .  |  .  .  .  |    |  .  .  .  |  .  .  .  |  .  .  .  |
 |  .  .  .  |  .  .  .  |  .  .  .  |    |  .  .  .  |  .  .  .  |  .  .  .  |
 |-----------+-----------+-----------|    |-----------+-----------+-----------|
 |  .  .  .  |  .  .  .  |  .  .  .  |    |  .  .  .  |  .  .  .  |  .  .  p  |
 | YZ  .  .  |  .  .  .  |  .  .  p  |    |  .  .  .  |  .  .  .  | YZ  .  p  |
 |  .  .  .  |  .  .  .  |  .  .  .  |    |  .  .  .  |  .  .  .  |  .  .  p  |
 +-----------------------------------+    +-----------------------------------+
 Other grids exist, but these will serve as common examples.

If candidate Z isn't present in any of the peer cells, then I'm wasting my time searching for an XY-Wing, XYZ-Wing, XY-Chain, or any other technique that connects the bivalue cells. If candidate Z is present in at least one of the peer cells, then it's easy enough to check for an XY-Wing or an XYZ-Wing. If neither wing is present, then I know that any XY-Chain connecting these two cells must link the X in the XZ cell to the Y in the YZ cell. This adds a constraint to the XY-Chain search that isn't present if you start with a random bivalue cell and search for a bivalue peer that contains at least one of its candidates.

Unfortunately, as in the following example, it's all too often that a puzzle comes to a screeching stop with many bivalue cells exposed. I know of no way to further reduce the multitude of search possibilities.

Code:
 XZ=r5c3, YZ=r7c1, Z=1  =>  r7c3<>1

 ... elimination would be on half of a bilocation on <1> in [r7] and [b7]
 ... it might prove productive, so look for a connection

 (1=4)r5c3 - (4=5)r4c3 - (5=3)r9c3 - (3=1)r7c1  =>  r7c3<>1
 +--------------------------------------------------------------+
 |  5     129   123   |  139   18    4     |  189   7     6     |
 |  134   8     6     |  2     7     39    |  149   159   345   |
 |  134   19    7     |  159   158   6     |  2489  129   23    |
 |--------------------+--------------------+--------------------|
 |  2     3    b45    |  8     9     1     |  7     6     45    |
 |  8     6    a14    |  7     3     5     |  1249  129   24    |
 |  7     15    9     |  6     4     2     |  13    35    8     |
 |--------------------+--------------------+--------------------|
 | d13    7     28-1  |  4     6     38    |  5     23    9     |
 |  6     25    58    |  159   15    389   |  23    4     7     |
 |  9     4    c35    |  35    2     7     |  6     8     1     |
 +--------------------------------------------------------------+
 # 58 eliminations remain

Regards, Danny
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Marty R.



Joined: 12 Feb 2006
Posts: 5179
Location: Rochester, NY, USA

PostPosted: Mon Dec 14, 2009 4:37 pm    Post subject: Reply with quote

Quote:
If you have two cells XY - YZ where X might be a pincer, start coloring on Z from YZ and see where that takes you?

I guess I don't understand why the coloring doesn't start with the Y from XY. That's what happened in Ted's chain, starting with the number that shared the cell with the pincer 6. And isn't that the usual thing with XY-Chains in general?
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storm_norm



Joined: 18 Oct 2007
Posts: 1741

PostPosted: Sun Dec 20, 2009 3:50 am    Post subject: Reply with quote

Quote:
Again, I find this approach of extending the vertex of a potential xy-wing easy to perform and surprisingly fruitful.

Ted


whatever works for you. its good to have some kind of search criteria when attacking a puzzle and it looks like this particular way is working for you.
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