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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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 Posted: Mon Mar 01, 2010 1:28 pm Post subject: |
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| 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 |
+------------+---------------+------------+
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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: 3355
Location: near Detroit, Michigan, USA
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| Posted: Mon Mar 01, 2010 6:31 pm Post subject: |
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| 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: 3355
Location: near Detroit, Michigan, USA
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| Posted: Mon Mar 01, 2010 6:41 pm Post subject: |
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| 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: 64
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| Posted: Mon Mar 01, 2010 10:08 pm Post subject: |
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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: 5770
Location: Rochester, NY, USA
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| Posted: Mon Mar 01, 2010 10:33 pm Post subject: |
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| 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: 3355
Location: near Detroit, Michigan, USA
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| Posted: Mon Mar 01, 2010 10:44 pm Post subject: |
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| 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: 5770
Location: Rochester, NY, USA
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| Posted: Mon Mar 01, 2010 10:50 pm Post subject: |
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| 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: 64
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| Posted: Tue Mar 02, 2010 12:19 am Post subject: |
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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: 5770
Location: Rochester, NY, USA
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| Posted: Mon Mar 22, 2010 1:55 am Post subject: |
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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: 3355
Location: near Detroit, Michigan, USA
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| Posted: Mon Mar 22, 2010 5:02 am Post subject: |
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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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