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M-Wing question
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Marty R.



Joined: 12 Feb 2006
Posts: 5162
Location: Rochester, NY, USA

PostPosted: Sun Feb 28, 2010 3:06 am    Post subject: M-Wing question Reply with quote

Code:

+--------------+-----------+--------------+
| 7   3   2456 | 469 12 46 | 125  159 8   |
| 25  15  9    | 7   3  8  | 125  6   4   |
| 24  148 2468 | 469 12 5  | 3    7   29  |
+--------------+-----------+--------------+
| 9   45  1    | 3   8  7  | 456  2   56  |
| 8   2   7    | 46  5  46 | 9    3   1   |
| 3   6   45   | 1   9  2  | 8    45  7   |
+--------------+-----------+--------------+
| 245 7   3    | 8   6  9  | 1245 145 25  |
| 6   9   58   | 2   4  1  | 7    58  3   |
| 1   48  248  | 5   7  3  | 246  489 269 |
+--------------+-----------+--------------+

Play this puzzle online at the Daily Sudoku site

This is the 2-28 VH which is solvable by an XY-Wing or UR. Note the 25 cells in boxes 19. Is that a valid M-Wing connected by 5's? The connection isn't the usual two simple strong links, but the effect seems to be the same, as if the extra 5's weren't present in r7c78.
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keith



Joined: 19 Sep 2005
Posts: 3174
Location: near Detroit, Michigan, USA

PostPosted: Sun Feb 28, 2010 3:33 am    Post subject: Reply with quote

Marty,

Code:
+----------------+----------------+----------------+
| 7    3    2456 | 469  12   46   | 125  159  8    |
| 25a  15   9    | 7    3    8    | 125c 6    4    |
| 24   148  2468 | 469  12   5    | 3    7   -29   |
+----------------+----------------+----------------+
| 9    45   1    | 3    8    7    | 456  2    56   |
| 8    2    7    | 46   5    46   | 9    3    1    |
| 3    6    45   | 1    9    2    | 8    45   7    |
+----------------+----------------+----------------+
| 245  7    3    | 8    6    9    |1-245 145  25b  |
| 6    9    58   | 2    4    1    | 7    58   3    |
| 1    48   248  | 5    7    3    |-246  489  269  |
+----------------+----------------+----------------+

Well, you have the basis of an M-wing.

In an astounding case of grouped coloring on 5, you can show that if a is 5, b is 5. Also, by a separate path, if b is 5, a is 5. So, a and b are the basis of an M-wing, which is completed by the strong link ac in R2.

Making the three eliminations shown.

By the way, you do not need to show if a is 5, b is 5. You only need the half wing, if b is 5 a is 5.

Logic:

b is 2, or
b is 5, a is 5, c is 2.
b and c are pincers on 2.

Keith Very Happy
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Marty R.



Joined: 12 Feb 2006
Posts: 5162
Location: Rochester, NY, USA

PostPosted: Sun Feb 28, 2010 4:51 am    Post subject: Reply with quote

Thanks Keith. Hopefully, my M-Wings can expand if I don't dismiss them because of the lack of two simple strong links and start looking more closely.

In this case, I established that if b=5, then a=5. However, I'm not seeing how b=5 if a=5.

Quote:
Logic:

b is 2, or
b is 5, a is 5, c is 2.
b and c are pincers on 2.

I think in different terms. If b=5, then r7c1<5>r2c7.
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Marty R.



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PostPosted: Sun Feb 28, 2010 4:52 am    Post subject: Reply with quote

Thanks Keith. Hopefully, my M-Wings can expand if I don't dismiss them because of the lack of two simple strong links and start looking more closely.

In this case, I established that if b=5, then a=5. However, I'm not seeing how b=5 if a=5.
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keith



Joined: 19 Sep 2005
Posts: 3174
Location: near Detroit, Michigan, USA

PostPosted: Sun Feb 28, 2010 5:23 am    Post subject: Reply with quote

Code:
+----------------+----------------+----------------+
| 7    3    2456 | 469  12   46   | 125f 159  8    |
| 25a  15   9    | 7    3    8    | 125c 6    4    |
| 24   148  2468 | 469  12   5    | 3    7   -29   |
+----------------+----------------+----------------+
| 9    45d  1    | 3    8    7    | 456  2    56   |
| 8    2    7    | 46   5    46   | 9    3    1    |
| 3    6    45   | 1    9    2    | 8    45e  7    |
+----------------+----------------+----------------+
| 245  7    3    | 8    6    9    |1-245 145  25b  |
| 6    9    58d  | 2    4    1    | 7    58   3    |
| 1    48   248  | 5    7    3    |-246  489  269  |
+----------------+----------------+----------------+

If a is 5 (true), so are the cells d. If d are true, so is e. If e is true, so is f. If f is true, so is b.

(As you go through this, you need to mark the cells that are not true, if a is true.)

Keith
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Marty R.



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PostPosted: Sun Feb 28, 2010 6:15 am    Post subject: Reply with quote

Thanks, I can follow it, but it's way too convoluted for me to have found it.
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tlanglet



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

PostPosted: Sun Feb 28, 2010 2:10 pm    Post subject: Reply with quote

Marty R. wrote:
Thanks, I can follow it, but it's way too convoluted for me to have found it.


Marty,

In many situations, I find that following implications is very similar to your original question about the extra 5s in r7c78 possibly influencing the link between the two 25s in r2c1 and r7c9.

In this specific case, I would walk the path as:
If r2c1=5, then r2c2<>5, then r4c2=5 due to the strong link in col2,
If r4c2=5, then r4c9<>5, then r7c9=5 due tot he strong link in col9.

Going the other direction, I get:
If r7c9=5, then r7c1<>5, then r2c1=5 due to the strong link in col1.

Viewed in this manner, I do not find the path as convoluted.

Ted
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daj95376



Joined: 23 Aug 2008
Posts: 3855

PostPosted: Sun Feb 28, 2010 3:34 pm    Post subject: Reply with quote

FWIW: Ted's cells are the vertices of a (222) Swordfish, with the base set being the three columns with strong links.

Code:
 Swordfish c129\r247  =>  r247c7,r7c8 <> 5
 +-----------------------------------+
 |  .  .  5  |  .  .  .  |  5  5  .  |
 | *5 *5  .  |  .  .  .  | -5  .  .  |
 |  .  .  .  |  .  .  5  |  .  .  .  |
 |-----------+-----------+-----------|
 |  . *5  .  |  .  .  .  | -5  . *5  |
 |  .  .  .  |  .  5  .  |  .  .  .  |
 |  .  .  5  |  .  .  .  |  .  5  .  |
 |-----------+-----------+-----------|
 | *5  .  .  |  .  .  .  | -5 -5 *5  |
 |  .  .  5  |  .  .  .  |  .  5  .  |
 |  .  .  .  |  5  .  .  |  .  .  .  |
 +-----------------------------------+

Since it's a (222) Swordfish, it can also be described as a continuous loop with the same eliminations.

Code:
(5): r2c2 = r4c2 - r4c9 = r7c9 - r7c1 = r2c1 - loop  =>  r247c7,r7c8 <> 5

Notice in the loop that r2c1, r4c2, and r7c9 are treated as true.
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Marty R.



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Posts: 5162
Location: Rochester, NY, USA

PostPosted: Sun Feb 28, 2010 4:31 pm    Post subject: Reply with quote

Quote:
In this specific case, I would walk the path as:
If r2c1=5, then r2c2<>5, then r4c2=5 due to the strong link in col2,
If r4c2=5, then r4c9<>5, then r7c9=5 due tot he strong link in col9.

Going the other direction, I get:
If r7c9=5, then r7c1<>5, then r2c1=5 due to the strong link in col1.

Viewed in this manner, I do not find the path as convoluted.

Ted, starting with r7c9, that's exactly my path. As to going the other way, it's really not convoluted; that was a poor choice of words. All I meant was that I probably wouldn't have thought to try that train of thought.

I have one (hopefully) final question for anybody. In my previous M-Wings I used only simple strong links so it could always be proven that one cell forced the other and vice versa and either cell could be extended by the other number to form the pincers.

In this actual case it was shown that if a=5, then b=5 and vice versa. However, if I can prove that if a=5, then b=5, but can't prove it going the other way, does that mean that only b can be extended to form pincers?
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daj95376



Joined: 23 Aug 2008
Posts: 3855

PostPosted: Sun Feb 28, 2010 4:38 pm    Post subject: Reply with quote

[Withdrew: Top part of post redundant.]

Marty: In your original description, you seemed fixated on the chain for <5> and ignored any chain on <2>. To me, it seemed that you missed the M-Wing completely. But, I've been wrong many other times.

[Edit: changed wording to better describe my interpretation -- which turned out to be wrong, again!]


Last edited by daj95376 on Sun Feb 28, 2010 6:30 pm; edited 2 times in total
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Marty R.



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PostPosted: Sun Feb 28, 2010 5:55 pm    Post subject: Reply with quote

Quote:
Marty: In your original description, you were fixated on the chain for <5> and completely ignored any chain on <2>. To me, it seemed that you missed the M-Wing completely. But, I've been wrong many other times.

Unlike you, I've never been wrong. Laughing

I don't know if I was "fixated" on 5, but that's the connection I saw and the extension of the 2 solved the puzzle in one step. I just wanted to make sure that my method of connection was valid, as I was used to forming M-Wings only via simple strong link connections.
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keith



Joined: 19 Sep 2005
Posts: 3174
Location: near Detroit, Michigan, USA

PostPosted: Sun Feb 28, 2010 9:55 pm    Post subject: Reply with quote

Quote:
In this actual case it was shown that if a=5, then b=5 and vice versa. However, if I can prove that if a=5, then b=5, but can't prove it going the other way, does that mean that only b can be extended to form pincers?


Marty, YES!!

And, there is a little more:

1. Suppose a is 25 and b is 25, and you can prove that 5 in a forces 5 in b. You can add the strong link in 2 to b to make the M-wing.

2. Suppose a is 25 and b is 2567, and you can prove that 5 in a forces 5 in b. You can add the strong link in 2 to b to make the M-wing.

Something like this: a(25) - b(25x) = c(2y)

5 in a forces 5 in b. Not 2 in b forces 2 in c. a and c are pincers in 2. (x and y can be any candidates.)

Case 1 above is a half M-wing. Case 2 is a generalized M-wing.

Remember to check whether a and c are in the same house (row, column, box). If they are, you have an M-cycle, and additional eliminations are possible.

Keith
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Marty R.



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PostPosted: Sun Feb 28, 2010 10:09 pm    Post subject: Reply with quote

Thank you!!! Exclamation

Quote:
Case 1 above is a half M-wing.

If you used one and posted your solution, would you list that step as a half M-Wing, as opposed to M-Wing?

Quote:
Case 2 is a generalized M-wing

I think this might be a 2nd definition I've seen of a gM-Wing.
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keith



Joined: 19 Sep 2005
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PostPosted: Sun Feb 28, 2010 10:25 pm    Post subject: Reply with quote

Quote:

If you used one and posted your solution, would you list that step as a half M-Wing, as opposed to M-Wing?


I would call it an M-wing, but then I always show the cells involved when I post a solution.

Quote:
I think this might be a 2nd definition I've seen of a gM-Wing.


Quote:
`When I use a word,' Humpty Dumpty said, in rather a scornful tone, `it means just what I choose it to mean -- neither more nor less.'


Keith
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Marty R.



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PostPosted: Mon Mar 01, 2010 12:04 am    Post subject: Reply with quote

I'm too old to have remembered anything about Humpty Dumpty, so I'll take your word on his wisdom.

Thanks again, I have no further questions for this witness.
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keith



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PostPosted: Mon Mar 01, 2010 3:23 am    Post subject: Reply with quote

Marty,

Nataraj' solution of Friday's Freep says it all. Incredible!

http://www.dailysudoku.com/sudoku/forums/viewtopic.php?t=4364

Keith
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Marty R.



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

PostPosted: Mon Mar 01, 2010 4:24 am    Post subject: Reply with quote

keith wrote:
Marty,

Nataraj' solution of Friday's Freep says it all. Incredible!

http://www.dailysudoku.com/sudoku/forums/viewtopic.php?t=4364

Keith

I saw that but don't understand one iota of it. The two cells are 17 and 159?
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keith



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PostPosted: Mon Mar 01, 2010 4:44 am    Post subject: Reply with quote

Marty,

Starting from the beginning: R8 is the only cell in C3 that can be 4.
Code:
+----------------------------+----------------------------+----------------------------+
| 7        25       1        | 46       2456     3        | 8        56       9        |
| 25689    2359     23569    | 14679    1245679  124567   | 2457     3567     246      |
| 2569     4        23569    | 679      25679    8        | 257      1        26       |
+----------------------------+----------------------------+----------------------------+
| 4        379      3679     | 13678    13678    167      | 179      2        5        |
| 256      2357     23567    | 134678   12345678 9        | 147      678      1468     |
| 1        8        25679    | 467      24567    24567    | 479      679      3        |
+----------------------------+----------------------------+----------------------------+
| 29       6        279      | 5        1789     17       | 3        4        128      |
| 259      12579    4        | 136789   136789   167      | 1259     589      128      |
| 3        159      8        | 2        149      14       | 6        59       7        |
+----------------------------+----------------------------+----------------------------+

1. R7C6 is 1. Or,
2. R7C6 is 7. In which case, R8C2 is 7 (and not 1). In which case, R9C2 in C2 is 1.

R7C6 and R9C2 are pincers on 1. Puzzle solved.

Keith
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Marty R.



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

PostPosted: Mon Mar 01, 2010 6:27 am    Post subject: Reply with quote

keith wrote:
Marty,

Starting from the beginning: R8 is the only cell in C3 that can be 4.
Code:
+----------------------------+----------------------------+----------------------------+
| 7        25       1        | 46       2456     3        | 8        56       9        |
| 25689    2359     23569    | 14679    1245679  124567   | 2457     3567     246      |
| 2569     4        23569    | 679      25679    8        | 257      1        26       |
+----------------------------+----------------------------+----------------------------+
| 4        379      3679     | 13678    13678    167      | 179      2        5        |
| 256      2357     23567    | 134678   12345678 9        | 147      678      1468     |
| 1        8        25679    | 467      24567    24567    | 479      679      3        |
+----------------------------+----------------------------+----------------------------+
| 29       6        279      | 5        1789     17       | 3        4        128      |
| 259      12579    4        | 136789   136789   167      | 1259     589      128      |
| 3        159      8        | 2        149      14       | 6        59       7        |
+----------------------------+----------------------------+----------------------------+

1. R7C6 is 1. Or,
2. R7C6 is 7. In which case, R8C2 is 7 (and not 1). In which case, R9C2 in C2 is 1.

R7C6 and R9C2 are pincers on 1. Puzzle solved.

Keith

OK, that's easy enough to follow. But what's M-Wingish about it? I'm assuming this is not trial and error. So what's there to tell the player to look at r7c6?
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strmckr



Joined: 18 Aug 2009
Posts: 61

PostPosted: Mon Mar 01, 2010 8:46 am    Post subject: Reply with quote

Quote:
1 r8c2 -1- r8c9 =1= r7c9 -1- r7c6 -7- r7c3 =7= r8c2 => r8c2<>1


losely described above doesn't match any of the m-wings which covers all variations.

but it does match a "s - wing" which i have covered here

i also did post a copy of both threads to this site in (about 4 thread topics down u can find them...)

(these are an aic chain of very short length always discontinuous )

Quote:

So what's there to tell the player to look at r7c6?


the bivalve digits 1 & 7


look for a strong link for digits "1" in a space.
&
look for a strong link for digits "7" in a space.


one cell from both 1 and 7 strongly linked cells must see the bivalve cell

if both 1 & 7 (end points) can see each other
then end points
1 <> 7 ,
7 <> 1

Code:

-------------------------------------------------
|  .   .   .  |  .    .   .    |   .    .    /  |
|  .   .   .  |  .    .   .    |   .    .    /  |
|  .   .   .  |  .    .   .    |   .    .    /  |
-------------------------------------------------
|  \   \   7  |  .   17   .    |   .    .    1  |
|  \  7-1  \  |  .    .   .    |   .    .   1-7 |
|  \   \   \  |  .    .   .    |   .    .    /  |
-------------------------------------------------
|  .   .   .  |  .    .   .    |   .    .    /  |
|  .   .   .  |  .    .   .    |   .    .    /  |
|  .   .   .  |  .    .   .    |   .    .    /  |
-------------------------------------------------
reference:
\ <> 7
/ <> 1
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