View previous topic :: View next topic 
Author 
Message 
Marty R.
Joined: 12 Feb 2006 Posts: 5256 Location: Rochester, NY, USA

Posted: Mon Mar 06, 2006 2:58 am Post subject: Unique rectangles 


I am spotting unique rectangles from time to time, but, unfortunately, none have been the exact kind I've seen in examples and my logical reasoning skills, or lack of same, are a problem. I have one similar to the Unique Rectangle Type 3 from the brain bashers site, shown below, but I don't understand the explanation. (Only cols. 1–3 shown).
What does he mean by a locked set and a pseudosingle square? If there's a locked set of <18> in col. 1, why is he removing the "1" from r5c1 rather than the "1" or "8" from r9c1?
Code:  
789 6 79
5 2 1 
4 78 3 

3 9 2 
17 5 4 
6 17 8 

1789 4 79
2 37 5 
18 138 6 
 
Unique Rectangles Type 3. There are almost two occurrences of a locked pair in <79>. However, we can never end up with all four squares only containing <79> otherwise we could swap the <7> & <9> and still have a valid solution, which is not allowed. Therefore, one of the values <1> or <8> must appear in one of R1C1 or R7C1, but we don't know which square, nor which number. But we can treat <18> as a pseudosingle square and look for a locked set containing it. We therefore have a locked set of <18> in C1 and can remove the <1> from R5C1. NOTE: the pseudosingle square can contain more than two additional numbers. 

Back to top 


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

Posted: Mon Mar 06, 2006 5:45 am Post subject: Unique rectangles 


Marty
I don’t understand the terms or unique rectangles but, for what it’s worth, the following comes to mind.
If you accept that 1 or 8 must be in r1c1 or r7c1, I think you can make the proposed elimination. If 1 occupies one of these cells it cannot enter r5c1. On the other hand if 8 occupies one of these cells, r9c1 contains 1 and again r5c1 cannot.
A locked set is a set with the same number of candidates as the order of the set. Here set means a subset of a column, a subset of a row or a subset of a box. You might have a set of two cells in one column which must contain 1 or 8. These values can then be eliminated from the other cells in the column.
In your fragment, either
 1 is in r1c1 or r7c1 (when there is a locked pair (18) in r19c1 or r79c1 or
 8 is in r1c1 or r7c1 (when there is a locked pair (18) in r19c1 or r79c1.
Whichever applies, 1 and 8 can be eliminated from all cells in column 1 except for r179c1. Here, of course, only the elimination of 1 is of any value.
Does this make any sense?
Steve 

Back to top 


keith
Joined: 19 Sep 2005 Posts: 3192 Location: near Detroit, Michigan, USA

Posted: Mon Mar 06, 2006 10:53 pm Post subject: Unique rectangles 


Marty,
It goes like this:
Explanation 1:
Code: 

789 6 79
5 2 1 
4 78 3 

3 9 2 
17 5 4 
6 17 8 

1789 4 79
2 37 5 
18 138 6 


The unique rectangle has the two <79>'s as the floor, and the roof squares are <1789> and <789>. The rectangle cannot have <79> at all four corners, for then the puzzle will have no unique solution. So, at least ONE of the roof squares is not <7> OR <9>. ONE of them is possibly <7> OR <9>. The other possibilites in the roof squares are <1> and <8>.
Notice the bottom left square, it is <18>, and forms a triple ("Locked Set?) with the roof squares: <1, 8, 7 OR 9>. <7 OR 9> (the "Pseudo Single"?) is one possibility. So, no other squares in C1 can be <1>, R5C1 must be 7, etc.
Explanation 2:
Look at it another way: If R5C1 is <1>, then the first column of the puzzle reduces to
Code: 

79 6 79
5 2 1 
4 78 3 

3 9 2 
1 5 4 
6 17 8 

79 4 79
2 37 5 
8 138 6 


which is not unique.
Explanation 3:
Maybe the following is clearer: At least one of the roof squares is not <7> OR <9>. However, <9> does not occur in any other squares in C1. So, one of the roof squares must be <9>, neither of them can be <7>, and R5C1 must be <7>.
The first explanation is a little more general; the third is, I think, easier to figure out.
Cute, isn't it?
Keith 

Back to top 


Marty R.
Joined: 12 Feb 2006 Posts: 5256 Location: Rochester, NY, USA

Posted: Tue Mar 07, 2006 1:36 am Post subject: 


Thanks Steve and Keith. I'd be lying if I said I could follow all that logic, but you both have given me some ideas as to approaches to these things that I wouldn't have otherwise realized. I'll try to apply them to the puzzle I'm working on, and if it doesn't work I'll be back. 

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
