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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Mon Mar 06, 2006 2:58 am Post subject: Unique rectangles |
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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 pseudo-single 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 |
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|3 9 2 |
|17 5 4 |
|6 17 8 |
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|1789 4 79|
|2 37 5 |
|18 138 6 |
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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 pseudo-single 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 pseudo-single square can contain more than two additional numbers. |
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Steve R
Joined: 24 Oct 2005 Posts: 289 Location: Birmingham, England
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Posted: Mon Mar 06, 2006 5:45 am Post subject: Unique rectangles |
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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 |
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Mon Mar 06, 2006 10:53 pm Post subject: Unique rectangles |
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Marty,
It goes like this:
Explanation 1:
Code: |
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|789 6 79|
|5 2 1 |
|4 78 3 |
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|3 9 2 |
|17 5 4 |
|6 17 8 |
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|1789 4 79|
|2 37 5 |
|18 138 6 |
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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: |
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|79 6 79|
|5 2 1 |
|4 78 3 |
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|3 9 2 |
|1 5 4 |
|6 17 8 |
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|79 4 79|
|2 37 5 |
|8 138 6 |
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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 |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Tue Mar 07, 2006 1:36 am Post subject: |
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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. |
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