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gM-Wing Type 6A Example

 
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Bud



Joined: 06 May 2010
Posts: 47
Location: Tampa, Florida

PostPosted: Tue Jun 08, 2010 3:57 pm    Post subject: gM-Wing Type 6A Example Reply with quote

I have found quite a few of Keith's original M-Wings, but after a long search I think I have finally found an example of a half M-Wing. I think the problem is it takes a while to learn how to look for them. The 4-cell pattern is marked with asterisks. Like the W-Wing, this is one of the simplest examples of a 2 digit AIC. This fact has made it easier for me to find it.

gM-Wing Type6A Example
Code:
 
 |-------------------+--------------------+--------------------|
 | 5789 34789  3579  |    6     2   349   | 3578     1    35   |
 |    1 34678 23567  |   78    48    34   |    9 34578   235   |
 | 2789 34789  2379  |  789     1     5   | 2378  3478     6   |
 |-------------------+--------------------+--------------------|
 |    4   679  1679  |    3    67     2   |  158   589   159   |
 |    3     5     8  |    4     9     1   |    6     2     7   |
 |  279  1679 12679  |    5    67     8   |   13*   39*    4   |
 |-------------------+--------------------+--------------------|
 |    6  3789  3579  |    1    48    49   | 2357 -3579  2359   |
 |  579   179     4  |    2     3     6   |  175*  579     8   |
 |   89     2    13  |   89     5     7   |    4     6    13*  |
 |-------------------+--------------------+--------------------|


The original puzzle is the daily tough puzzle of Jan. 4,2010 at Sudoku.com.au. An 1 X-color contradiction is the only advanced technique I used to reach this point in the puzzle. Edit: I have used a 1 ER beginning at r8c2 to eliminate a 1 from r4c2.


Last edited by Bud on Wed Jun 09, 2010 5:44 pm; edited 1 time in total
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daj95376



Joined: 23 Aug 2008
Posts: 3854

PostPosted: Tue Jun 08, 2010 5:36 pm    Post subject: Reply with quote

r4c7 contains <1> and ruins any chance of a gM-Wing that I see using your cells.
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Bud



Joined: 06 May 2010
Posts: 47
Location: Tampa, Florida

PostPosted: Wed Jun 09, 2010 6:27 pm    Post subject: Reply with quote

You are right DAJ. As I originally presented this, it is was an almost Half M-Wing but it is actually true. I have added a 1 ER move to eliminate a 1 from r4c2 as shown below to show this. Now (1)r6c7-(1)r6c23=(1)r4c3-(1)r9c3=(1)r9c9. Thus it is a Half M-Wing since a 1 in r6c7 gives a 1 in r9c9 independent of the 1 in r4c7.

gM-Wing Type6A Example
Code:
 
 |-------------------+--------------------+--------------------|
 | 5789 34789  3579  |    6     2   349   | 3578     1    35   |
 |    1 34678 23567  |   78    48    34   |    9 34578   235   |
 | 2789 34789  2379  |  789     1     5   | 2378  3478     6   |
 |-------------------+--------------------+--------------------|
 |    4   679  1679  |    3    67     2   |  158   589   159   |
 |    3     5     8  |    4     9     1   |    6     2     7   |
 |  279  1679 12679  |    5    67     8   |   13*   39*    4   |
 |-------------------+--------------------+--------------------|
 |    6  3789  3579  |    1    48    49   | 2357 -3579  2359   |
 |  579   179     4  |    2     3     6   |  175*  579     8   |
 |   89     2    13  |   89     5     7   |    4     6    13*  |
 |-------------------+--------------------+--------------------|
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daj95376



Joined: 23 Aug 2008
Posts: 3854

PostPosted: Wed Jun 09, 2010 8:08 pm    Post subject: Reply with quote

Bud wrote:
You are right DAJ. As I originally presented this, it is was an almost Half M-Wing but it is actually true. I have added a 1 ER move to eliminate a 1 from r4c2 as shown below to show this. Now (1)r6c7-(1)r6c23=(1)r4c3-(1)r9c3=(1)r9c9. Thus it is a Half M-Wing since a 1 in r6c7 gives a 1 in r9c9 independent of the 1 in r4c7.

I see a lot of effort creating a useless chain segment that could be replaced with (1)r6c7 - r8c7 = (1)r9c9 without using the ER elimination. I still don't see anything remotely resembling a gM-Wing.
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