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Nataraj HS Dec 1, VH

 
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keith



Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
PostPosted: Wed Dec 01, 2010 2:42 pm    Post subject: Nataraj HS Dec 1, VH Reply with quote
Code:
+-------+-------+-------+
| . . . | 9 . 5 | . . 4 |
| . 2 . | . . . | 5 . . |
| 8 . 3 | . 4 7 | . . 9 |
+-------+-------+-------+
| 3 6 4 | . . 9 | . . . |
| . . . | . . . | . . . |
| . . . | 5 . . | 4 6 7 |
+-------+-------+-------+
| 1 . . | 8 6 . | 9 . 5 |
| . . 5 | . . . | . 2 . |
| 9 . . | 7 . 2 | . . . |
+-------+-------+-------+
After basics:
Code:
+----------------+----------------+----------------+
| 7    1    6    | 9    23   5    | 23   8    4    |
| 4    2    9    | 136  138  1368 | 5    7    36   |
| 8    5    3    | 26   4    7    | 26   1    9    |
+----------------+----------------+----------------+
| 3    6    4    | 12   7    9    | 8    5    12   |
| 5    8    7    | 46   12   46   | 13   9    123  |
| 2    9    1    | 5    38   38   | 4    6    7    |
+----------------+----------------+----------------+
| 1    7    2    | 8    6    34   | 9    34   5    |
| 6    34   5    | 134  9    134  | 7    2    8    |
| 9    34   8    | 7    5    2    | 16   34   16   |
+----------------+----------------+----------------+

edit: The following is not correct. See the later messages in this thread. - Keith
Quote:
Note the 38 UR which has a pseudocell 16 in R2B2. This forms an XY-wing with 26 12 and eliminates 1 in R5C5.

There is also a 34 DP in R789 which solves R8C6 as 1.
Keith

Last edited by keith on Wed Dec 01, 2010 8:28 pm; edited 1 time in total
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nataraj



Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
PostPosted: Wed Dec 01, 2010 3:37 pm    Post subject: Reply with quote
Wow !

I am always fascinated by the hidden treasures that the sharp eyes of this community can pull out of even humble "VH"s Smile

All I saw was a boring xy-wing 123 ...
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keith



Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
PostPosted: Wed Dec 01, 2010 3:45 pm    Post subject: Reply with quote
nataraj wrote:
Wow !

I am always fascinated by the hidden treasures that the sharp eyes of this community can pull out of even humble "VH"s Smile

All I saw was a boring xy-wing 123 ...

My solution is easier because I did not have to find the box / column intersection that reduces R5C7 to 13 Laughing

Keith
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peterj



Joined: 26 Mar 2010
Posts: 974
Location: London, UK
PostPosted: Wed Dec 01, 2010 5:52 pm    Post subject: Reply with quote
Just to show his this forum can really go completely over the top on finding alternative solutions to simple ones!
Here is a move that can be played after only singles and results in only singles! Does a simpler move exist with the same property? (Does anyone care!)
Code:

After singles...
 *--------------------------------------------------------------------*
 | 7      1      6      | 9      23     5      | (23)   8      4      |
 | 4      2      9      | 136    138    1368   | 5      7      36     |
 | 8      5      3      | (2)6   4      7      | (2)6   1      9      |
 |----------------------+----------------------+----------------------|
 | 3      6      4      | 1-2    7      9      | 8      5      1(2)   |
 | 5      8      7      | 12346  123    1346   | 12(3)  9      1(2)(3)|
 | 2      9      1      | 5      38     38     | 4      6      7      |
 |----------------------+----------------------+----------------------|
 | 1      7      2      | 8      6      34     | 9      34     5      |
 | 6      34     5      | 134    9      134    | 7      2      8      |
 | 9      34     8      | 7      5      2      | 136    34     136    |
 *--------------------------------------------------------------------*
gm-wing(23) ; (2=3)r1c7 - r5c7=(3-2)r5c9=r4c9
with transport (2)r1c7 - r3c7=r3c4 ; r4c4<>2

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peterj



Joined: 26 Mar 2010
Posts: 974
Location: London, UK
PostPosted: Wed Dec 01, 2010 6:00 pm    Post subject: Re: Nataraj HS Dec 1, VH Reply with quote
keith wrote:
Note the 38 UR which has a pseudocell 16 in R2B2.

Keith, if the pseudocell is in row2 (as opposed to column5) I don't see how it can make the elimination of the 1 in r5c5? The DP might be prevented by r2c6=1 in the case where 6 is excluded?
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
PostPosted: Wed Dec 01, 2010 6:22 pm    Post subject: Reply with quote
I need a pseudo-cell education because I don't see what a 16 cell does.
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ronk



Joined: 07 May 2006
Posts: 398
PostPosted: Wed Dec 01, 2010 7:37 pm    Post subject: Re: Nataraj HS Dec 1, VH Reply with quote
peterj wrote:
keith wrote:
Note the 38 UR which has a pseudocell 16 in R2B2.

Keith, if the pseudocell is in row2 (as opposed to column5) I don't see how it can make the elimination of the 1 in r5c5?

Likely should be r2c4<>1.
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keith



Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
PostPosted: Wed Dec 01, 2010 8:26 pm    Post subject: Re: Nataraj HS Dec 1, VH Reply with quote
ronk wrote:
peterj wrote:
keith wrote:
Note the 38 UR which has a pseudocell 16 in R2B2.

Keith, if the pseudocell is in row2 (as opposed to column5) I don't see how it can make the elimination of the 1 in r5c5?

Likely should be r2c4<>1.

peterj is correct: I made a mistake. However, maybe I can skate by:
Code:
+----------------+----------------+----------------+
| 7    1    6    | 9    23   5    | 23   8    4    |
| 4    2    9    | 136# 138a 1368b| 5    7    36   |
| 8    5    3    | 26c  4    7    | 26   1    9    |
+----------------+----------------+----------------+
| 3    6    4    | 12d  7    9    | 8    5    12   |
| 5    8    7    | 46   12@  46   | 13   9    123  |
| 2    9    1    | 5    38   38   | 4    6    7    |
+----------------+----------------+----------------+
| 1    7    2    | 8    6    34   | 9    34   5    |
| 6    34   5    | 134  9    134$ | 7    2    8    |
| 9    34   8    | 7    5    2    | 16   34   16   |
+----------------+----------------+----------------+

The pseudocell is 16 in ab. In making a chain (or an elimination) a "6" has to see b, but a "1" has to see both a and b.

The XY-wing is (ab)cd and it makes only ronk's elimination of 1 in R2C4, #. My elimination in R5C5, @, is incorrect (but lucky!).

However, I also pointed out the DP in B789, which solves R8B6, $, as 1. The combination of $ is 1 and # is not 1 solves the puzzle.

Maybe we can call this technique: Using a pseudo-cell to get a pseudo-solution. Embarassed

Keith
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