| View previous topic :: View next topic |
| Author |
Message |
Bud
Joined: 06 May 2010
Posts: 47
Location: Tampa, Florida
|
 Posted: Tue Jun 08, 2010 3:57 pm Post subject: gM-Wing Type 6A Example |
 |
|
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 |
|
 |
daj95376
Joined: 23 Aug 2008
Posts: 3854
|
| Posted: Tue Jun 08, 2010 5:36 pm Post subject: |
|
|
| r4c7 contains <1> and ruins any chance of a gM-Wing that I see using your cells. |
|
| Back to top |
|
|
Bud
Joined: 06 May 2010
Posts: 47
Location: Tampa, Florida
|
| Posted: Wed Jun 09, 2010 6:27 pm Post subject: |
|
|
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 |
|
|
daj95376
Joined: 23 Aug 2008
Posts: 3854
|
| Posted: Wed Jun 09, 2010 8:08 pm Post subject: |
|
|
| 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 |
|
|
|
|
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
|
FAQ
Search
Memberlist
Usergroups
Register
Profile
Log in to check your private messages
Log in
