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VH+ by gsf (game 0057)

 
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ravel



Joined: 21 Apr 2006
Posts: 536

PostPosted: Sat Nov 01, 2008 11:47 am    Post subject: VH+ by gsf (game 0057) Reply with quote

Code:
. . 7 8 . . . . .
. 2 . . . 1 6 . .
3 . . . . . . 9 .
9 . . 5 . . . 8 .
. . . . 2 . . . .
. 1 . . . 4 . . 2
. 4 . . . . . . 1
. . 9 3 . . . . .
. . . . . 7 2 . . gsf

Maybe quickly done with an xy-chain, but i solved it with four uniqueness moves.
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daj95376



Joined: 23 Aug 2008
Posts: 3855

PostPosted: Sat Nov 01, 2008 5:50 pm    Post subject: Reply with quote

M-Wing:

Code:
 [r5c9]<>6 [r5c9]=3 [r2c9]<>3 [r2c8]=3 [r9c8]<>3 [r9c3]=3 [r9c3]<>6 [r6c3]=6  =>  [r5c1],[r6c8]<>6
 +--------------------------------------------------------------+
 |  4     9     7     |  8     6     3     |  1     2     5     |
 |  58    2     58    |  4     9     1     |  6    #37   #37    |
 |  3     6     1     |  2     7     5     |  48    9     48    |
 |--------------------+--------------------+--------------------|
 |  9     3     2     |  5     1     6     |  47    8     47    |
 |  7-6   5     4     |  79    2     8     |  39    1     36*   |
 |  678   1    +68*   |  79    3     4     |  59    5-6   2     |
 |--------------------+--------------------+--------------------|
 |  2     4     35    |  6     58    9     |  3578  357   1     |
 |  1     7     9     |  3     458   2     |  58    456   68    |
 |  56    8    @356+  |  1     45    7     |  2    @345   9     |
 +--------------------------------------------------------------+
__________________________________________________________________________________________________
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keith



Joined: 19 Sep 2005
Posts: 3179
Location: near Detroit, Michigan, USA

PostPosted: Sat Nov 01, 2008 7:42 pm    Post subject: Reply with quote

daj95376 wrote:
M-Wing:

Danny,

Wow! I am impressed!

Keith
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Marty R.



Joined: 12 Feb 2006
Posts: 5166
Location: Rochester, NY, USA

PostPosted: Sat Nov 01, 2008 9:42 pm    Post subject: Reply with quote

keith wrote:
daj95376 wrote:
M-Wing:

Danny,

Wow! I am impressed!

Keith

I don't understand enough to be impressed. I can certainly see where the eliminations occur based on r5c9 <> 6, but what leads up to that step? I see a variety of symbols in the grid, but I don't know where the M-Wing starts. Question
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keith



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

PostPosted: Sat Nov 01, 2008 10:11 pm    Post subject: Reply with quote

Marty,

Look here for my explanation:

http://www.dailysudoku.com/sudoku/forums/viewtopic.php?t=2143

Keith
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Marty R.



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

PostPosted: Sun Nov 02, 2008 12:30 am    Post subject: Reply with quote

Keith, I've referred to that thread numerous times and after looking at it again, I still have the same questions. I know, I'm dense. Embarassed
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daj95376



Joined: 23 Aug 2008
Posts: 3855

PostPosted: Sun Nov 02, 2008 2:08 am    Post subject: Reply with quote

Marty,

My implication chain starts with cell [r5c9]. That might be a good place for you to start.

The (*) cells mark the start and end of a conjugate relationship on (6).

The (#) cells mark a strong link on (3) in [r2].

The (@) cells mark a strong link on (3) in [r9].

The (+) cells mark a strong link on (6) in [c3].

In keith's subsequent thread, he also indicates a grouped strong link on (3) in [c8]. Neither one of us mentioned the presence of a strong link on (3) in [c9].

If it wasn't for the presence of (3) in cell [r7c8] and (5) in [r9c3], this would be a perfect example of a 6-cell M-Wing.

===== ===== ===== To Add Confusion

IF I had written my chain using NL notation, it would have been an AIC with [r7c8]<>3 and [r9c3]<>5 being additional eliminations Exclamation
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Marty R.



Joined: 12 Feb 2006
Posts: 5166
Location: Rochester, NY, USA

PostPosted: Sun Nov 02, 2008 4:10 am    Post subject: Reply with quote

Quote:
If it wasn't for the presence of (3) in cell [r7c8] and (5) in [r9c3], this would be a perfect example of a 6-cell M-Wing.

Maybe that explains things. My knowledge of M-Wings doesn't go beyond the basics of two strong links connecting two identical bivalue cells.

Thanks.
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Asellus



Joined: 05 Jun 2007
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Location: Sonoma County, CA, USA

PostPosted: Sun Nov 02, 2008 10:14 am    Post subject: Reply with quote

Marty R. wrote:
Maybe that explains things. My knowledge of M-Wings doesn't go beyond the basics of two strong links connecting two identical bivalue cells.

Let me give it a try.

Yes, M-Wings, especially in their "generalized" form, are not as easy to spot as W-Wings. However, they have a very similar structure, which, if understood, might help with spotting them.

The W-Wing always has an identical pair of non-peer bivalues, each of which "sees" the opposite ends of an external strong inference link in one of their digits. If the two bivalues are XY and the external strong inference is in Y, then the Xs in the bivalue cells are the pincers of the W-Wing. Schematically, it can be written as
(XY) - Y=Y - (YX)
where the bivalue cells are shown in parentheses.

"Y=Y" can be anything with a strong inference: a simple conjugate pair, or a grouped conjugate pair, or the pincers of a wing, or the pincers of a wing with one or more of them "transported," etc.

Usually, with the W-Wing, we focus first on those non-peer matching bivalues, then look for the external strong link. I could also write those bivalues with strong link symbols inside the cell parentheses:
(X=Y) - Y=Y - (Y=X)
The strong link symbols inside the parentheses make it clear that these must be bivalue cells.

For an M-Wing, it helps to turn things on their head (much as an "M" is a "W" on its head!) and look first for that (external) strong inference link which, we will assume, is otherwise useless... the "Y=Y" part. If one end of that strong link can see a bivalue cell, (XY), and the other end shares its cell with X, then you might have an M-Wing. All you need is a conjugate link (or other strong link, to be really general!) from that X in the shared cell to another X that serves as a useful pincer with the X in the (XY) bivalue cell. (Read that again slowly, if necessary, to make sure you've got it.)

In Danny's M-Wing, the external strong link is formed by the otherwise useless "backwards" Skyscraper in rows 2 and 9. This gives us pincer <3>s (the "Y=Y" part) in r2c9 and r9c3, or:
(3)r2c9=(3)r9c3.

[This comes from the "useless Skyscraper" structure:
(3)r2c9=(3)r2c8 - (3)r9c8=(3)r9c3
which can be collapsed to just:
(3)r2c9=(3)r9c3.]

The <3> in r2c9 can "see" the (36) bivalue in r5c9. And, the <3> in r9c3 shares its cell with a <6>. So, we have most of the makings of an M-Wing! All we need to complete it is a <6> that is conjugate (or otherwise strongly linked) to that <6> in r9c3. We have two choices: the <6> at r6c3 or the <6> at r9c1. Either one will do since they both have a common peer with the <6> in that (36) bivalue at r5c9. That creates the M-Wing pincer <6>s that eliminate <6> from r5c1.

The generalized form of the M-Wing is, thus:
(XY) - Y = (Y-X) = X
instead of the W-Wing:
(XY) - Y = Y - (YX)

The weak link notation in parenthese, (Y-X), indicates that there can be other candidates in that cell... it does not need to be a bivalue because the link is weak. If we make all of the links explicit, then we have...

M-Wing:
(X=Y) - Y=(Y-X)=X
W-Wing:
(X=Y) - Y=Y - (Y=X)

This is just a short, symbolic way of expressing what I described above. The key is seeing that "Y=Y" part that is at the heart of both techniques.
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nataraj



Joined: 03 Aug 2007
Posts: 1029
Location: Vienna, Austria

PostPosted: Sun Nov 02, 2008 10:18 am    Post subject: Re: VH+ by gsf (game 0057) Reply with quote

ravel wrote:

Maybe quickly done with an xy-chain, but i solved it with four uniqueness moves.


I took the hint from ravel, and remembering Asellus' astute way of looking at URs as strong links, made use of the 58-68-56 possible DP in the leftmost tower:

Code:

+--------------------------+--------------------------+--------------------------+
| 4       9       7        | 8       6       3        | 1       2       5        |
| 58      2       58       | 4       9       1        | 6       37      37       |
| 3       6       1        | 2       7       5        | 48      9       48       |
+--------------------------+--------------------------+--------------------------+
| 9       3       2        | 5       1       6        | 47      8       47       |
| 67      5       4        | 79      2       8        | 39      1       36       |
| 678     1       68       | 79      3       4        | 59      56      2        |
+--------------------------+--------------------------+--------------------------+
| 2       4       35       | 6       58      9        | 3578    357     1        |
| 1       7       9        | 3       458     2        | 58      456     68       |
| 56      8       356      | 1       45      7        | 2       345     9        |
+--------------------------+--------------------------+--------------------------+

To break the DP, r6c1=7 or r9c3=3, so why not "transport" one of the ends?
If r6c1=7 then r5c1=6 r5c9=3. Which means at least one of r5c9,r9c3 must be 3. Insert this new strong link into the existing SL-structure for "3":

and immediately solve r7c3=5. This cracks the puzzle.

---

edited to change wording (and typo) from "Which means either r5c9=3 or r9c3=3" to "at least one of"


Last edited by nataraj on Sun Nov 02, 2008 10:36 am; edited 1 time in total
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nataraj



Joined: 03 Aug 2007
Posts: 1029
Location: Vienna, Austria

PostPosted: Sun Nov 02, 2008 10:27 am    Post subject: Reply with quote

OMG. I have successfully avoided looking at the thread so far (in order not to prejudice my solution). Isn't it funny that for once I venture into a bit more intricate UR eliminations and everyone else is on generalized m-wings? Laughing Laughing Laughing

I'll try and digest the m loop soon as I find some time. Sounds terriffic!
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ravel



Joined: 21 Apr 2006
Posts: 536

PostPosted: Sun Nov 02, 2008 11:07 am    Post subject: Reply with quote

Nice solutions.

In mine no chains/transports are needed, but a type 6 (?) UR, leading to a type 4 UR, strong links in the 6-cell DP above, and a 2-cell BUG to finish it off.
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daj95376



Joined: 23 Aug 2008
Posts: 3855

PostPosted: Sun Nov 02, 2008 11:25 am    Post subject: Reply with quote

Marty,

I owe you an apology Exclamation

I'm so fixated on W-Wings and M-Wings as chains -- i.e., the generalized forms -- that I often overlook some of the other constraints associated with the primary definitions. That happened this time with my ignoring the (3) in [r7c8] and the (5) in [r9c3].

As an AIC chain in NL notation, this M-Wing would look like

Code:
6- [r5c9] -3- [r1c9] =3= [r1c8] -3- [r9c8] =3= [r9c3] =6= [r6c3] ...
 => [r5c1],[r6c8]<>6, [r7c8]<>3, [r9c3]<>5

I just saw the string of links in (3) with a (6) on each end.

My apologies to you Exclamation

Danny
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nataraj



Joined: 03 Aug 2007
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Location: Vienna, Austria

PostPosted: Sun Nov 02, 2008 11:55 am    Post subject: Reply with quote

ravel wrote:
Nice solutions.

In mine no chains/transports are needed, but a type 6 (?) UR, leading to a type 4 UR, strong links in the 6-cell DP above, and a 2-cell BUG to finish it off.


Ah! I saw that (58) type 6 but it did not seem to do much (r7c7,r8c5<>8).
I failed to see the (45) type 4 (which removes 5 from r89c8)

That takes us to the 58-68-56 DP in a reduced grid:
Code:

+--------------------------+--------------------------+--------------------------+
| 4       9       7        | 8       6       3        | 1       2       5        |
| 58      2       58       | 4       9       1        | 6       37      37       |
| 3       6       1        | 2       7       5        | 48      9       48       |
+--------------------------+--------------------------+--------------------------+
| 9       3       2        | 5       1       6        | 47      8       47       |
| 67      5       4        | 79      2       8        | 39      1       36       |
| 678     1       68       | 79      3       4        | 59      56      2        |
+--------------------------+--------------------------+--------------------------+
| 2       4       35       | 6       8       9        | 357     357     1        |
| 1       7       9        | 3       45      2        | 58      46      68       |
| 56      8       356      | 1       45      7        | 2       34      9        |
+--------------------------+--------------------------+--------------------------+

I'll look for the weasels after lunch. Thanks, ravel!
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cgordon



Joined: 04 May 2007
Posts: 769
Location: ontario, canada

PostPosted: Sun Nov 02, 2008 12:29 pm    Post subject: Reply with quote

In my recent searches for ERís, Iíve been noticing cases
where one of the strong link cells does not actually go through
a hinge but instead leaves an x-wing or some other available solution.

Here, the bottom <8> in C6 removes R7C3.
The top <8> in C6 leaves an x-wing that also removes R7C3.
Does this have a name? If not I want naming rights!


Code:
            
+-------+-------+-------+   
| . . . | . . . | . . . |   
| 8 . 8 | . . . | . . . |   
| . . . | . . . | 8 . 8 |   
+-------+-------+-------+   
| . . . | . . . | . . . |   
| 8 8 . | . . 8 | . . . |   
| 8 . 8 | . 8 . | . . . |   
+-------+-------+-------+   
| . . 8 | . 8 8 | 8 . . |   
| . 8 . | . 8 . | 8 . 8 |   
| 8 8 8 | . 8 . | . . 8 |
+-------+-------+-------+
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Marty R.



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PostPosted: Sun Nov 02, 2008 4:43 pm    Post subject: Reply with quote

Thanks Asellus and Danny, and to the latter, no apologies necessary.
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daj95376



Joined: 23 Aug 2008
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PostPosted: Sun Nov 02, 2008 5:05 pm    Post subject: Reply with quote

Craig: You're (almost) describing ...

Code:
 Kraken X-Wing r26\c13 w/ext fin [r6c5]  =>  [r7c3]<>8

        X-Wing r26\c13                   =>  [r7c3]<>8
 [r6c5]=8 => [r5c6]<>8 => [r7c6]=8       =>  [r7c3]<>8
 +-----------------------------------+
 |  .  .  .  |  8  .  .  |  .  .  .  |
 | *8  . *8  |  .  .  .  |  .  .  .  |
 |  .  .  .  |  .  .  .  |  8  .  8  |
 |-----------+-----------+-----------|
 |  .  .  .  |  .  .  .  |  .  8  .  |
 |  8  8  .  |  .  .  8  |  .  .  .  |
 | *8  . *8  |  . #8  .  |  .  .  .  |
 |-----------+-----------+-----------|
 |  .  . -8  |  .  8  8  |  8  .  .  |
 |  .  8  .  |  .  8  .  |  8  .  8  |
 |  8  8  8  |  .  8  .  |  .  .  8  |
 +-----------------------------------+

It's also the simpler:

Code:
 Sashimi X-Wing c26\r57 w/int fins [r89c2]  =>  [r7c3]<>8
 +-----------------------------------+
 |  .  .  .  |  8  .  .  |  .  .  .  |
 |  8  .  8  |  .  .  .  |  .  .  .  |
 |  .  .  .  |  .  .  .  |  8  .  8  |
 |-----------+-----------+-----------|
 |  .  .  .  |  .  .  .  |  .  8  .  |
 |  8 *8  .  |  .  . *8  |  .  .  .  |
 |  8  .  8  |  .  8  .  |  .  .  .  |
 |-----------+-----------+-----------|
 |  .  . -8  |  .  8 *8  |  8  .  .  |
 |  . #8  .  |  .  8  .  |  8  .  8  |
 |  8 #8  8  |  .  8  .  |  .  .  8  |
 +-----------------------------------+

In the second example, another way of viewing it is that the top (8) in [c6] exposes a Locked Candidate 2 in [b7] that performs the elimination.
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Asellus



Joined: 05 Jun 2007
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PostPosted: Sun Nov 02, 2008 10:15 pm    Post subject: Reply with quote

daj95376 wrote:
IF I had written my chain using NL notation, it would have been an AIC with [r7c8]<>3 and [r9c3]<>5 being additional eliminations

Those eliminations would require that this AIC be a continuous loop, and no matter how I look at it, I can't see it. In fact, r7c8=3 in the solution! So, there'd better be no such loop.
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