dailysudoku.com Forum Index dailysudoku.com
Discussion of Daily Sudoku puzzles
 
 FAQFAQ   SearchSearch   MemberlistMemberlist   UsergroupsUsergroups   RegisterRegister 
 ProfileProfile   Log in to check your private messagesLog in to check your private messages   Log inLog in 

gM-Wing Type 6A Example

 
Post new topic   Reply to topic    dailysudoku.com Forum Index -> Solving techniques, and terminology
View previous topic :: View next topic  
Author Message
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
Back to top
View user's profile Send private message
daj95376



Joined: 23 Aug 2008
Posts: 3855

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.
Back to top
View user's profile Send private message
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*  |
 |-------------------+--------------------+--------------------|
Back to top
View user's profile Send private message
daj95376



Joined: 23 Aug 2008
Posts: 3855

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.
Back to top
View user's profile Send private message
Display posts from previous:   
Post new topic   Reply to topic    dailysudoku.com Forum Index -> Solving techniques, and terminology All times are GMT
Page 1 of 1

 
Jump to:  
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum


Powered by phpBB © 2001, 2005 phpBB Group