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M-Wing question
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Marty R.



Joined: 12 Feb 2006
Posts: 5123
Location: Rochester, NY, USA

PostPosted: Mon Mar 01, 2010 1:28 pm    Post subject: Reply with quote

Quote:
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

OK, thanks. I can see where that matches the definition of a gM-Wing that Danny gave me.

Does this qualify as some variation of an M-Wing?

Code:

+------------+---------------+------------+
| 7    12 9  | 256 156  126  | 3   4    8 |
| 26   8  16 | 3   14   124  | 79  79   5 |
| 5    3  4  | 9   8    7    | 6   2    1 |
+------------+---------------+------------+
| 269  4  7  | 1   69   2689 | 5   689  3 |
| 369  5  16 | 48  7    3489 | 189 1689 2 |
| 2369 12 8  | 26  369  5    | 179 1679 4 |
+------------+---------------+------------+
| 4    9  5  | 7   13   138  | 2   18   6 |
| 8    6  2  | 45  1459 149  | 14  3    7 |
| 1    7  3  | 468 2    468  | 48  5    9 |
+------------+---------------+------------+

Play this puzzle online at the Daily Sudoku site

If r9c6=8, then r5c4=8. Can the 4 be extended from r5c4 to r5c6 to form pincers with r9c6?
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keith



Joined: 19 Sep 2005
Posts: 3150
Location: near Detroit, Michigan, USA

PostPosted: Mon Mar 01, 2010 6:31 pm    Post subject: Reply with quote

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


Marty,

The original recipe for M-wings involved finding two bivalue cells and looking for how they might be connected. That is a variant on how to find W-wings, so I still use it.

Here is a different recipe:

Choose any cell, a, that has only two candidates, XY. See if you can find another cell, b, such that
1) a = X forces b = X.
2) b contains Y as a candidate.

If you can find a third cell, c, such that bc is a strong link in Y, then ac are the pincers of an M-wing, and eliminate Y.

Keith
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keith



Joined: 19 Sep 2005
Posts: 3150
Location: near Detroit, Michigan, USA

PostPosted: Mon Mar 01, 2010 6:41 pm    Post subject: Reply with quote

Marty R. wrote:

Does this qualify as some variation of an M-Wing?

Code:

+------------+---------------+------------+
| 7    12 9  | 256 156  126  | 3   4    8 |
| 26   8  16 | 3   14   124  | 79  79   5 |
| 5    3  4  | 9   8    7    | 6   2    1 |
+------------+---------------+------------+
| 269  4  7  | 1   69   2689 | 5   689  3 |
| 369  5  16 | 48b 7    3489c| 189 1689 2 |
| 2369 12 8  | 26  369  5    | 179 1679 4 |
+------------+---------------+------------+
| 4    9  5  | 7   13   138  | 2   18   6 |
| 8    6  2  |-45  1459 149  | 14d 3    7 |
| 1    7  3  | 468 2   -468  | 48a 5    9 |
+------------+---------------+------------+

Play this puzzle online at the Daily Sudoku site

If r9c6=8, then r5c4=8. Can the 4 be extended from r5c4 to r5c6 to form pincers with r9c6?

Yes, except for the typo:If r9c6=8, ... should be If r9c7=8, ...

8 in a forces 8 in b, a and c are pincers in 4. This is a half-wing.

Note also that 8 in b forces 8 in a. a and d are pincers in 4.

I suppose, this is another half. But, note that the logic a => b is not the reverse of b => a.

Keith
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strmckr



Joined: 18 Aug 2009
Posts: 61

PostPosted: Mon Mar 01, 2010 10:08 pm    Post subject: Reply with quote

there is a few moves from those cells

Code:
.--------------.-----------------.--------------.
| 7     12  9  | 256  156   126  | 3    4     8 |
| 26    8   16 | 3    14    124  | 79   79    5 |
| 5     3   4  | 9    8     7    | 6    2     1 |
:--------------+-----------------+--------------:
| 269   4   7  | 1    69    2689 | 5    689   3 |
| 369   5   16 | 48@  7     3489 | 189@ 1689  2 |
| 2369  12  8  | 26   369   5    | 179  1679  4 |
:--------------+-----------------+--------------:
| 4     9   5  | 7    13    138  | 2    18    6 |
| 8     6   2  | 5-4  1459  149  | 14@   3    7 |
| 1     7   3  | 468  2     468  | 48@   5    9 |
'--------------'-----------------'--------------'


m-wing (type 1a)
4 r8c4 -4- r8c7 =4= r9c7 =8= r5c7 -8- r5c4 -4- r8c4 => r8c4<>4


& another m-wing (type 1b) i found is:

Code:
+------------+---------------+------------+
| 7    12 9  | 256 156  126  | 3   4    8 |
| 26   8  16 | 3   14   124  | 79  79   5 |
| 5    3  4  | 9   8    7    | 6   2    1 |
+------------+---------------+------------+
| 269  4  7  | 1   69   2689 | 5   689  3 |
| 369  5  16 | 48@ 7    3489@| 189 1689 2 |
| 2369 12 8  | 26  369  5    | 179 1679 4 |
+------------+---------------+------------+
| 4    9  5  | 7   13   138  | 2   18   6 |
| 8    6  2  | 45  1459 149  | 14  3    7 |
| 1    7  3  | 468@ 2  68-4  | 48@ 5    9 |
+------------+---------------+------------+


4 r9c6 -4- r9c7 -8- r9c4 =8= r5c4 =4= r5c6 -4- r9c6 => r9c6<>4


Last edited by strmckr on Mon Mar 01, 2010 10:42 pm; edited 2 times in total
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Marty R.



Joined: 12 Feb 2006
Posts: 5123
Location: Rochester, NY, USA

PostPosted: Mon Mar 01, 2010 10:33 pm    Post subject: Reply with quote

Quote:
s-wing
4 r8c4 -4- r8c7 =4= r9c7 =8= r5c7 -8- r5c4 -4- r8c4 => r8c4<>4

Thanks, I'll have to research that, having never heard of an S-Wing.

Quote:
Yes, except for the typo:If r9c6=8, ... should be If r9c7=8, ...

It wasn't a typo.

Quote:
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.

I was wondering if it was one of these, which you referred to as a generalized M-Wing. I think the real question I have is in these cases, does it only work one way, that the bivalue has to prove the polyvalue, thus the extension of the other number can be made only from the polyvalue cell?

Thanks to both of you.
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keith



Joined: 19 Sep 2005
Posts: 3150
Location: near Detroit, Michigan, USA

PostPosted: Mon Mar 01, 2010 10:44 pm    Post subject: Reply with quote

Quote:
that the bivalue has to prove the polyvalue, thus the extension of the other number can be made only from the polyvalue cell?
Absolutely, yes!

Keith
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Marty R.



Joined: 12 Feb 2006
Posts: 5123
Location: Rochester, NY, USA

PostPosted: Mon Mar 01, 2010 10:50 pm    Post subject: Reply with quote

keith wrote:
Quote:
that the bivalue has to prove the polyvalue, thus the extension of the other number can be made only from the polyvalue cell?
Absolutely, yes!

Keith

That's why I was getting invalid solutions. I played it by extending the 48 after proving it was 8 based on the 468 cell and taking out the rest of the 4s in c6. Finally, I think I now know the rules.
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strmckr



Joined: 18 Aug 2009
Posts: 61

PostPosted: Tue Mar 02, 2010 12:19 am    Post subject: Reply with quote

well it seems i miss understood which locations you where using..

if u consider both digits are in box 7 then you have a m-wing.

(what i shown early was for an s-wing)

hopefully it doesn't cause too much confusion or compound it more..

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       49-1      4-1    | 6     59    7    |
'---------------------'--------------------------'------------------'


1- r9c2 =1= r8c2 =7= r7c3 -7- r7c6 -1 => r9c56<>1
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Marty R.



Joined: 12 Feb 2006
Posts: 5123
Location: Rochester, NY, USA

PostPosted: Mon Mar 22, 2010 1:55 am    Post subject: Reply with quote

I think I found an opportunity for a generalized M-Wing per Keith's definition. This is a Vanhegan Fiendish:

Code:

+-------+-------+-------+
| . 7 . | . 1 . | . 5 . |
| 5 . . | 3 2 4 | . . 6 |
| . . 6 | . . . | 9 . . |
+-------+-------+-------+
| 2 . . | 4 3 1 | . . 8 |
| . . . | 9 . 2 | . . . |
| 1 . . | 5 7 6 | . . 4 |
+-------+-------+-------+
| . . 2 | . . . | 4 . . |
| 8 . . | 1 4 7 | . . 9 |
| . 4 . | . 9 . | . 8 . |
+-------+-------+-------+

Play this puzzle online at the Daily Sudoku site

After basics:

Code:

+------------+--------+------------+
| 4  7   8   | 6 1 9  | 3   5   2  |
| 5  19  19  | 3 2 4  | 8   7   6  |
| 3  2   6   | 7 5 8  | 9   4   1  |
+------------+--------+------------+
| 2  569 579 | 4 3 1  | 57  69  8  |
| 67 356 4   | 9 8 2  | 157 136 35 |
| 1  8   39  | 5 7 6  | 2   39  4  |
+------------+--------+------------+
| 9  15  2   | 8 6 35 | 4   13  7  |
| 8  356 35  | 1 4 7  | 56  2   9  |
| 67 4   17  | 2 9 35 | 16  8   35 |
+------------+--------+------------+

Play this puzzle online at the Daily Sudoku site

The first time I did it it was pencil and paper; I think I have the same grid. Couldn't find a move. But then I noticed that 1 in r9c7 proves 1 in r5c8. Extend the 6 from r5c8 to r4c8 and you have flightlessness. But then transport the 6 from r9c7 to r5c1 and the puzzle solves.

I know Keith mentioned that in this type of Wing, the other number can be extended only from the polyvalue cell. But once that's done it would seem that either pincer can be transported. If not, then I made a lucky mistake.
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keith



Joined: 19 Sep 2005
Posts: 3150
Location: near Detroit, Michigan, USA

PostPosted: Mon Mar 22, 2010 5:02 am    Post subject: Reply with quote

Marty,

Yes, you are correct. The M-wing has pincers on 6 at R4C8 and R9C7. The latter can be extended to R5C1, eliminating 6 in R4C2 and R5C8.

Once you have established the pincers, the usual rules on extensions apply. They are no different than for the pincers of any other chain.

Keith
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