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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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 Posted: Fri May 20, 2011 9:04 am Post subject: Another tough Menneske |
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Looks like this one is interesting. The challenge is to solve it with short chains.
I did find a Sue de Coq! I have not yet completed it.
| Code: | Puzzle: M6306219sh(37)
+-------+-------+-------+
| 1 8 . | . . . | . 4 9 |
| . 9 4 | . . 1 | 6 7 . |
| . . . | . . 5 | . . . |
+-------+-------+-------+
| . 2 9 | . . 3 | . . . |
| . . . | 1 . 8 | . . . |
| . . . | 5 . . | 3 8 . |
+-------+-------+-------+
| . . . | 7 . . | . . . |
| . 4 1 | 2 . . | 8 9 . |
| 9 5 . | . . . | . 2 6 |
+-------+-------+-------+ |
Keith |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Fri May 20, 2011 8:04 pm Post subject: |
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I don't know whether the criteria stretches to a simple als move?
| Quote: | | xy-wing-als(35-2) (2=38)r2c45 - (3=5)r2c9 - (5=2)r1c7 ; r1c56<>2 |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat May 21, 2011 5:05 am Post subject: |
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| Quote: | | I did find a Sue de Coq! |
Ms. de Coq and I are not acquainted.
M-Wing (25). The 5 in r1c7 proves 5 in r2c1; r1c56<>2. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat May 21, 2011 1:57 pm Post subject: |
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| Marty R. wrote: | | Ms. de Coq and I are not acquainted. |
I would hope not. I believe it translates as "a boy named Sue".
After basics: | Code: | +----------------+----------------+----------------+
| 1 8 235 | 6 237 27 | 25 4 9 |
| 235 9 4 | 38 238 1 | 6 7 35 |
| 2367 367 2367 | 9 4 5 | 12 13 8 |
+----------------+----------------+----------------+
| 8 2 9 | 4 6 3 | 157 15 1-57 |
| 3457 37 357 | 1 27 8 | 9 6 24 |
| 467 1 67 | 5 9 27 | 3 8 24 |
+----------------+----------------+----------------+
| 236 36 2368 | 7 1358 9 | 4 -135 135 |
| 37 4 1 | 2 35 6 | 8 9 357 |
| 9 5 378 | 38 138 4 | 17 2 6 |
+----------------+----------------+----------------+ |
Sue de Coq: Take a look at B9C9.
The possibilities in R78C9 are 1357, but they cannot be 1 and 7 because of R9C7, nor 3 and 5 because of R2C9.
So, the solution in R78C9 is 1 or 7, and 3 or 5. Making the eliminations shown.
In other words, R78C9 makes two pseudocells, 17 and 35. I am still trying to understand this thing. The available explanations are not very good. In fact, if you read the discussion on the Sudopedia article, you will see that someone thinks the Sudopedia definition is wrong.
http://www.sudopedia.org/wiki/Talk:Sue_de_Coq
Here is the original article:
http://forum.enjoysudoku.com/two-sector-disjoint-subsets-t2033.html
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat May 21, 2011 4:55 pm Post subject: |
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Thank you. I will try and digest that stuff, but don't hold your breath waiting for me to discover one. 
| Quote: | | So, the solution in R78C9 is 1 or 7, and 3 or 5. |
Is that the same as saying that the combinations can be 13, 15, 37 or 57? |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat May 21, 2011 5:39 pm Post subject: |
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| Marty R. wrote: | | Quote: | | So, the solution in R78C9 is 1 or 7, and 3 or 5. |
Is that the same as saying that the combinations can be 13, 15, 37 or 57? |
Yes.
Keith |
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