View previous topic :: View next topic 
Author 
Message 
RogerC
Joined: 08 Oct 2005 Posts: 14 Location: High Wycombe, Bucks, England

Posted: Fri Apr 07, 2006 5:18 pm Post subject: Daily Squiggly Sudoku: Tue 4Apr2006 Really very hard 


Code:  3 0 7 0 0 0 9 2 1
0 2 9 0 0 0 8 0 0
0 9 0 2 0 0 5 0 0
0 0 0 0 9 0 0 0 0
6 0 0 0 5 0 0 9 7
9 0 0 0 7 0 0 0 2
2 0 8 0 0 0 0 5 0
0 0 2 0 0 0 0 1 0
0 4 0 0 2 0 6 0 9 
As you can see, I haven't got far with this extremely difficult squiggley!
Every other time I have asked for help, I have missed something fairly obvious, but I have spent a lot of time on this one and I'm beginning to think it requires a solving technique that I just don't know!
Any help would be very welcome. 

Back to top 


mastertsai
Joined: 08 Apr 2006 Posts: 2


Back to top 


Steve R
Joined: 24 Oct 2005 Posts: 289 Location: Birmingham, England

Posted: Sat Apr 08, 2006 12:10 pm Post subject: The blind leading the blind 


Roger
I am new to squigglies and this one drove me to distraction, mostly, but not entirely, at the position you posted.
You will have spotted the pair in box 7: (79) in r78c4.
Now consider where 1 and 3 might be placed in columns 1, 2 and 3. Neither can go in box 2 so they must fall in boxes 1, 4 and 7. In particular, the 1 and 3 of box 7 must lie somewhere in columns 1, 2 and 3. However, the 5 is constrained to these columns as well by the 5 given in box 9 and the known pair (79). In short box 7 has a pair (79) in column 4 and a triple (135) at the bottom left (r9c1, r8c2 and r9c3).
As a result r7c5 contains 6. Other 6s follow.
I hope this makes sense.
Steve 

Back to top 


RogerC
Joined: 08 Oct 2005 Posts: 14 Location: High Wycombe, Bucks, England

Posted: Sat Apr 08, 2006 4:24 pm Post subject: 


Thanks for your reply, Steve, but I'm afraid I don't understand any of it!
Firstly the 79 pair in r78c4 is not an exclusive pair as the 7 is possible in r8c2 and r9c1 in box 7, and r9c4 in column 4. In addition, r8c2 has a possible 6, so the 135 'triple' is not valid.
I have never read a book or even an internet article on sudokus so my technique could be said to be amateur to say the least, but this is the first one that has beaten me. Perhaps there is an essential trick that I need to learn for this one?
Roger 

Back to top 


Steve R
Joined: 24 Oct 2005 Posts: 289 Location: Birmingham, England

Posted: Sat Apr 08, 2006 8:13 pm Post subject: 


Is not 7 excluded from r8c2 and r9c1 by the facts that box 1 can admit 7 only in column 1 or column 2 and that box 4 is similarly constrained?
Steve 

Back to top 


RogerC
Joined: 08 Oct 2005 Posts: 14 Location: High Wycombe, Bucks, England

Posted: Sun Apr 09, 2006 10:49 am Post subject: 


Of course! Why is the answer always so simple?!
Thank you very much. Sorry to be so thick!
Roger 

Back to top 


SL Guest

Posted: Tue Apr 11, 2006 2:38 am Post subject: 


Thank you for Steve to give a great contribution to solve the "7".
However, I still don't understand why triple (1,3,5) appear in box 7, as in col 2, 6 may be appeared in r1c2 or r8c2, so how to eliminate the 6 for r8c2?
Please give your sincere explanation again.
Thanks. 

Back to top 


Steve R
Joined: 24 Oct 2005 Posts: 289 Location: Birmingham, England

Posted: Tue Apr 11, 2006 3:10 pm Post subject: The blind leading the blind 


I had hoped that an expert at squigglies would post a straightforward answer to this puzzle. Let me try again.
We are, I think, agreed that r78c4 contain the pair (79). Call r9c1, r8c2 and r9c3 the “triad” for short. The object is not really to exclude 6 from the triad; it is to force 1, 3 and 5 into it.
The case of 5. Bearing the pair (79) in mind, the 5 given in r7c8 forces 5 into the triad.
The case of 3. Columns 2 and 3 must each contain a 3. Box 2 cannot house either of them; nor can box 1. They must fall in boxes 4 and 7. That is, the 3 of box 7 is forced into the triad.
The case of 1 involves two steps. First consider where 1 is to be placed in columns 7 and 9. Boxes 8 and 9 are not available, leaving just box 3 and box 6. The 1 which goes in box 3 column7 or box 3 column 9 occupies row 1. Now consider where 1 is to be placed in columns 1, 2 and 3. Box 2 cannot house one of these 1s since it already has a 1 in the first row. These three 1s must occupy boxes 1, 4 and 7. That in box 7 is forced into the triad.
Steve 

Back to top 


SL Guest

Posted: Tue Apr 11, 2006 3:56 pm Post subject: 


Steve,
You have very strong analytical mind.
Thank you for your explanation,
but it still a long journey to solve it. 

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
