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sideli123
Joined: 20 Dec 2005 Posts: 7

Posted: Sat Feb 04, 2006 4:41 pm Post subject: OK I've got a tuff one 


I've been solving evil and very hard puzzles now for months and havent been stumped. I find the doubles and triples but this puzzle has me scratching my head. Can someone help. It starts with this
 1 48
9  2
53 7 
 5 96
3 9 1
89 4 
 6 85
2  6
54 7 
I got it to this and can go no further
7 1 48
918  2
453 7 619
4 5 96
35 9 481
89 4 
71 6 854
82  6
546 7 
Thanks for the help 

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keith
Joined: 19 Sep 2005 Posts: 3184 Location: near Detroit, Michigan, USA

Posted: Sat Feb 04, 2006 7:18 pm Post subject: A route to the solution 


There is a pair <26> in R1.
There is a triple <238> in C5.
There is a pair <45> in C5.
One you recognize these, then R6C5 is <6>, R2C4 is <6>, and the rest involves only pinned squares.
I hope this helps.
Keith 

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David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Posted: Sat Feb 04, 2006 7:29 pm Post subject: This puzzle has two solutions 


Hi, sidelit23! It's nice to hear from you.
You can break through the logjam at this position (below) by using a "doubleimplication chain." I've only shown the candidate lists that matter for the sake of this argument.
Code:    7  1   4 8
9 1 8    2  
4 5 3 7 28 28 6 1 9
  4   5 9 6 
3  5 26 9  4 8 1
 8 9 4 236    
7 39 1 239 23 6 8 5 4
8  2      6
5 4 6  7     
Let's examine what happens if we set r3c5 = 2. Then
A. r3c5 = 2 ==> r7c5 = 3 ==> r7c2 = 9
B. (r3c5 = 2 & r7c5 = 3) ==> r6c5 = 6 ==> r5c4 = 2 ==> r7c4 = 9
So r3c5 cannot be a "2", and it must be an "8". This value is enough to let you (almost) complete the puzzle. dcb
PS When I got close to the end of this one I ended up with a "swordfish" pattern that could be completed in two different ways. So I think this puzzle has multiple solutions. Oh  I went back and checked more closely, and I found 6 distinct solutions. You can either find a "swordfish" with {3, 9} pairs or, by making a different choice, you can wind up with two independent "nonunique rectangles", one in {3, 9} and the other in {5, 7}. 

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David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Posted: Sat Feb 04, 2006 7:54 pm Post subject: The "uniquity" assumption 


Keith wrote:  Once you recognize these, then R6C5 is <6>, R2C4 is <6>, and the rest involves only pinned squares. ... 
Hi, Keith!
We've had a discussion before about the "nonunique rectangles," and how one can sometimes solve a puzzle by recognizing them, or the closely related "BUG" pattern.
This puzzle is a pretty good example of the logical flaw in those methods. If you continue working through this you'll find a "nonunique rectangle" in r2c7, r2c9, r6c7, & r6c9. And if you make the assumption that the solution is unique you'll obtain a "3" at r2c7.
The only thing is, the solution you'll arrive at when you do this is not unique, because there are six cells in which "3" and "9" are interchangeable.
On top of that, if you assume that r2c7 <> 3 you'll arrive at a different set of solutions, with two embedded "nonunique rectangles."
Just thought you might be interested  this is an edifying example. dcb 

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David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Posted: Sat Feb 04, 2006 8:53 pm Post subject: Oops  I miscopied the puzzle 


Oops!
I just realized that I made a mistake when I copied this puzzle  I inadvertently put the "2" that should be in r2c7 in r2c8 instead. That's why I found multiple solutions.
I still think it (this puzzle with the "2" misplaced) is an interesting example. dcb 

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