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Free Press Feb 11, 2010 (Thursday)

 
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keith



Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA

PostPosted: Sat Feb 13, 2010 7:24 pm    Post subject: Free Press Feb 11, 2010 (Thursday) Reply with quote

Quite nice.
Code:
Puzzle: FP021110
+-------+-------+-------+
| . 8 . | 4 1 2 | . . . |
| . . 9 | . 7 . | 6 2 . |
| . 3 . | . . . | . . . |
+-------+-------+-------+
| . 6 . | . . . | . 7 . |
| . . 7 | 9 6 8 | 4 . . |
| . 1 . | . . . | . 3 . |
+-------+-------+-------+
| . . . | . . . | . 6 . |
| . 9 4 | . 3 . | 1 . . |
| . . . | 2 8 4 | . 9 . |
+-------+-------+-------+
Keith
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Sat Feb 13, 2010 8:44 pm    Post subject: Reply with quote

Quote:
R3c3 was forced = 2 based on any way to kill a 67-57-56 DP in the middle stack
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keith



Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA

PostPosted: Sat Feb 13, 2010 9:16 pm    Post subject: Reply with quote

Marty,

I don't see how your DP works.
After basics:
Code:
+----------------+----------------+----------------+
| 7    8    6    | 4    1    2    | 39   5    39   |
| 15   4    9    | 358  7    35   | 6    2    18   |
| 125  3    25   | 568  59   569  | 78   4    178  |
+----------------+----------------+----------------+
| 49   6    258  | 13   24   13   | 2589 7    2589 |
| 3    25   7    | 9    6    8    | 4    1    25   |
| 49   1    28   | 57   24   57   | 289  3    6    |
+----------------+----------------+----------------+
| 8    257  3    | 157  59   1579 | 257  6    4    |
| 25   9    4    | 567  3    567  | 1    8    257  |
| 6    57   1    | 2    8    4    | 357  9    357  |
+----------------+----------------+----------------+

To destroy the DP, there is a 5 in R8B8, or an 89 in R3B2C46? The DP is, I assume, 57 in R6B5, 67 in R8B8, and 56 in R3B2.

Quote:
I used a skyscraper on 2 in R57.


Keith


Last edited by keith on Sat Feb 13, 2010 9:29 pm; edited 1 time in total
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arkietech



Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA

PostPosted: Sat Feb 13, 2010 9:23 pm    Post subject: Reply with quote

I used a:
Quote:
skyscraper 2
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Sat Feb 13, 2010 10:36 pm    Post subject: Reply with quote

Quote:
To destroy the DP, there is a 5 in R8B8, or an 89 in R3B2C46? The DP is, I assume, 57 in R6B5, 67 in R8B8, and 56 in R3B2.

I could've screwed up twice (first on paper), but I just used your grid on Draw/Play and got the same result. What is there that you're contesting?

P.S. I never got as far as checking for X-Wings, which is where I'd have found the skyscraper.
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keith



Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA

PostPosted: Sat Feb 13, 2010 11:07 pm    Post subject: Reply with quote

Marty,

I'm not contesting anything, I'm asking for an explanation. Here's how I see it:
Code:
+----------------+----------------+----------------+
| 7    8    6    | 4    1    2    | 39   5    39   |
| 15   4    9    | 358  7    35   | 6    2    18   |
| 125  3    25   | 568  59   569  | 78   4    178  |
+----------------+----------------+----------------+
| 49   6    258  | 13   24   13   | 2589 7    2589 |
| 3    25   7    | 9    6    8    | 4    1    25   |
| 49   1    28   | 57   24   57   | 289  3    6    |
+----------------+----------------+----------------+
| 8    257  3    | 157  59   1579 | 257  6    4    |
| 25   9    4    | 567  3    567  | 1    8    257  |
| 6    57   1    | 2    8    4    | 357  9    357  |
+----------------+----------------+----------------+


The DP to be avoided is:

56 in R3 C46
57 in R6 C46
67 in R8 C46.

OK?

To avoid the DP:

One of R8 C46 is 5. I agree, that forces 2 in R3C3.

Or:

R3 C46 are a pseudocell 89 in R3B2. I do not see how that forces anything useful.

Keith
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Sun Feb 14, 2010 12:22 am    Post subject: Reply with quote

Quote:
R3 C46 are a pseudocell 89 in R3B2. I do not see how that forces anything useful.

Keith,

If r3c6=9, then it's pretty obvious how the 2 is forced.

If r3c4=8, I tried again and forced the 2. I'm sorry, I can't explain it, but it's a long and convoluted chain.

But once you get the 9 in r3c5, then the 5 in r7c5, it's easy going.
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