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Ema Nymton
Joined: 17 Apr 2009 Posts: 87

Posted: Sat Jun 07, 2014 10:33 pm Post subject: Stumped, Again ... 


.
Is this solvable?
Code: 
++++
 9 8 46  3 7 5  1 46 2 
 2 7 146  69 14 8  5 3 49 
 3 5 146  69 14 2  8 469 7 
++++
 5 6 3  2 9 14  7 14 8 
 7 14 8  5 6 3  9 2 14 
 14 2 9  7 8 14  3 5 6 
++++
 8 19 2  4 5 7  6 19 3 
 14 149 5  8 3 6  2 7 19 
 6 3 7  1 2 9  4 8 5 
++++

Play this puzzle online at the Daily Sudoku site
If it solvable, what is the step I am overlooking?
Thanks for any help you may be able to provide.
Ema Nymton
~ @ : o ?
. 

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arkietech
Joined: 31 Jul 2008 Posts: 1752 Location: Northwest Arkansas USA

Posted: Sat Jun 07, 2014 11:24 pm Post subject: 


Code: 
**
 9 8 46  3 7 5  1 46 2 
 2 7 146  69 14 8  5 3 49 
 3 5 146  69 14 2  8 469 7 
++
 5 6 3  2 9 14  7 14 8 
 7 *14 8  5 6 3  9 2 *14 
*14 2 9  7 8 14  3 5 6 
++
 8 19 2  4 5 7  6 19 3 
*14 149 5  8 3 6  2 7 91 
 6 3 7  1 2 9  4 8 5 
**
The 14 pairs shown form a Remote Pair removing 1 from r8c9 and solving the puzzle. 


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Ema Nymton
Joined: 17 Apr 2009 Posts: 87

Posted: Sat Jun 07, 2014 11:42 pm Post subject: 


.
Thank you. Impressive solution.
Ema Nymton
~ @ : o ?
. 

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dongrave
Joined: 06 Mar 2014 Posts: 242

Posted: Sun Jun 08, 2014 2:37 pm Post subject: 


I think you can also remove a bunch of the 1's if you 'color' them all in rows 4 through 8. I don't know what the technical solution name is for this (yet) but maybe the experts could tell us. Instead, I applied what's called a proof by contradiction in math terms (i.e. assume that any one the squares r4c8, r5c2, r6c6, r7c2, r8c1, or r8c9 contains a 1 , then chain them until you arrive at a contradiction). What do they call this in Sudoku solving terms  or is this generally not an accepted approach? Thanks, Don. 

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dongrave
Joined: 06 Mar 2014 Posts: 242

Posted: Sat Jun 14, 2014 6:29 pm Post subject: Hey! Coloring chains is cool! 


I just went back and read more about coloring and I now see how it works! Cool! I can see that in this puzzle if you color the 1's beginning at r7c9 (I used yellow) and then color r5c9 with red, then r5c2 with yellow, r6c1 with red, then r8c1 must be yellow  but it can't be yellow! Cool! I also see now the major difference between this and the 'assume and find contraction' approach in that the 'contradiction' approach truly is trial and error  and it seems to me that one should never have to resort to trial and error to solve a puzzle. I'd appreciate any thoughts. Oh, and please let me know if I'm wrong about the coloring that I've described above. I've been wrong before  so much so that I was thinking of asking for someone to add a 'very embarrassed' Emoticon (like maybe a bag over the head)! Thanks, Don. 

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Marty R.
Joined: 12 Feb 2006 Posts: 5470 Location: Rochester, NY, USA

Posted: Sat Jun 14, 2014 9:59 pm Post subject: Re: Hey! Coloring chains is cool! 


dongrave wrote:  I just went back and read more about coloring and I now see how it works! Cool! I can see that in this puzzle if you color the 1's beginning at r7c9 (I used yellow) and then color r5c9 with red, then r5c2 with yellow, r6c1 with red, then r8c1 must be yellow  but it can't be yellow! Cool! I also see now the major difference between this and the 'assume and find contraction' approach in that the 'contradiction' approach truly is trial and error  and it seems to me that one should never have to resort to trial and error to solve a puzzle. I'd appreciate any thoughts. Oh, and please let me know if I'm wrong about the coloring that I've described above. I've been wrong before  so much so that I was thinking of asking for someone to add a 'very embarrassed' Emoticon (like maybe a bag over the head)! Thanks, Don. 
Don, I assume you had a typo and started with r8c9. Coloring got its name from using colors, but it isn't necessary. In a Simple Coloring chain, you use a series of strong links, such as exists with the 1s. Using your four cells, you can say TFTF, or PlusMinusPlusMinus, just to use two examples.
The important consideration is that the chain has an even number of cells, thus the start and end of the chain have opposite "polarity" and any cell seeing both the start and end cells cannot contain the value since one or the other must be true. 

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