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istvan New kid on the Grid
Joined: 10 Nov 2006 Posts: 4 Location: Budapest, Hungary

Posted: Fri Nov 10, 2006 10:16 pm Post subject: Mirrorreflex in today's (Nov 10 2006) nightmare 


In today's nightmare there are several possibilities of using chains of empty rectangles. For example after few steps there are two such chains:
Code:  . . .* * .* . *
. 3 17 * 59 8 *
. 7 4. . .6 1 .
++
* 8 .1 . 35 2 *
* * .. 5 .* . .
. 5 .9 . 4. 7 .
++
. 9 85 . .2 4 .
* 1 72 * 83 6 .
* * .* * .. . .

The asterisks show the candidates for digit 4. The first mirrorreflex chain starts with R4C9 (that has a strong link with R4C1). The line of sight goes from R4C9 up into box 3, that has an empty rectangle. The ER reflects the beam to the left into row 1, where it is mirrored again in box 2 into column 5 because there is another ER in box 2. The beam is mirrored again in box 8 (a third ER in that box) to the left into R9, and reaches R9C1 that can see the other end of the original strong link at R4C1. Therefore digit 4 can be eliminated in R9C1.
Another mirrorreflex beam starts at R8C5 (that has a strong link with R8C1). The beam goes up in C5 to box 2 where it is mirrored by the ER to R1 into box 3, mirrored down along C9 into box 6, mirrored to the left in R5, and finally reaches R5C1 that can see the other end (R8C1) of the strong link as well. So digit 4 can be eliminated from R5C1.
I don't know whether this technique has already got a name; I call it mirrorreflex, because the several reflections of the line of sight are like the reflections in a mirrorreflex camera.
I think this technique is very easy to spot when one tries to find empty rectangles. 

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istvan New kid on the Grid
Joined: 10 Nov 2006 Posts: 4 Location: Budapest, Hungary

Posted: Sat Nov 11, 2006 11:54 pm Post subject: NonER reflection in mirrorreflex 


I found a different use of this technique. In the below example it works with more general reflections, not only with empty rectangles. (This arrangement is a random unfair sudoku, generated with SudoCue, after several steps)
Code:  * * .2 * 7* * 3
* * .* * .6 2 .
2 3 .6 * 84 * .
++
7 6 .8 2 91 . .
* * .3 7 52 8 6
. 2 84 1 6. 7 9
++
. 7 95 . 2. 6 .
. 5 2* * .7 * .
8 . .7 * .* * 2

The asterisks are the candidates for digit 9. R2C4 and R8C4 have a strong link. The beam goes from R2C4 to the left into box 1. Box 1 is not an ER because it doesn't have 4 empty cells in a rectangular form. It has a row of 3 cells (in row 3) that don't contain a candidate for digit 9. (R1C3 and R2C3 doesn't contain a candidate either, but nothing would change if they did.) This arrangement reflects the beam 180 degrees back into row 1. This is easy to explain: when R2C4 is true, all other candidates in R2 are eliminated, so one of the R1 cells of box 1 must be true (as R3 is empty). Effectively the line of sight for R2C4 is reflected into row 1.
The continuation of the mirrorreflex is similar to the yesterday arrangement: the beam goes to box 3 that is an ER (with a T shaped hinge), so the beam is reflected to C8 and reaches R8C8. Thus R8C8 sees both ends of the strong link (R8C4 directly, R2C4 in the reflex) and can be eliminated.
By the way, the beam may turn on to the left into R9 because box 9 is an ER, too, so it reaches R9C5, that this way sees both ends of the strong link and can be eliminated, too.
Does anyone think this mirrorreflex technique is the same as another already documented one? 

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Ruud Site Owner
Joined: 30 Dec 2005 Posts: 601

Posted: Sun Nov 12, 2006 5:39 pm Post subject: 


Hi istvan, welcome to the forum.
Taking a closer look at your initial example:
Code:  . . .*  .* . 
. 3 17 X 59 8 X
. 7 4. . .6 1 .
++
X 8 .1 . 35 2 X
 * .. 5 .* . .
. 5 .9 . 4. 7 .
++
. 9 85 . .2 4 .
X 1 72 X 83 6 .
 * .*  .. . . 
The candidates marked with X form a Swordfish pattern, which eliminates the candidates marked .
cheers,
Ruud _________________ “If the human brain were so simple that we could understand it, we would be so simple that we couldn't.”  Emerson M Pugh 

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Ron Moore Addict
Joined: 13 Aug 2006 Posts: 72 Location: New Mexico

Posted: Sun Nov 12, 2006 7:17 pm Post subject: 


istvan,
I'm not the most authoritative forum member to answer you, but take this for what it's worth.
It's happened to me many times  when I make successive eliminations of the same digit based on empty rectangles, in retrospect I often find that there is a single fish structure that would have yielded the same results, and sometimes more. I also noted the swordfish pattern in your first example, which Ruud mentions.
In your second example, reproduced below for convenience, the eliminations you mention are direct results of the empty rectangles in boxes 8 and 9. Both patterns use the conjugate (aka stronglylinked) pair in r3c58. Alternatively, you can view the eliminations as coming from finned Xwings.
Code:  * * .2 * 7* * 3
* * .* * .6 2 .
2 3 .6 * 84 * .
++
7 6 .8 2 91 . .
* * .3 7 52 8 6
. 2 84 1 6. 7 9
++
. 7 95 . 2. 6 .
. 5 2* * .7 * .
8 . .7 * .* * 2 
As to your general question, though, your technique can be useful. I would consider it as a usage of Alternating Inference Chains (AIC's), of the "grouped" variety. As you probably know, if there exists a chain of cells which are successively linked with alternating strong and weak links, and in which the first and last links are strong links, then a strong link exists between the endpoints of the chain. An inference can then be drawn in any cell which sees both endpoints. (Of course, the inference which can be drawn is not always useful, if it provides no new information, but sometimes it is.) The idea can be extended to chains in which the nodes are not necessarily single cells, but groups of two or three aligned cells within a box. For a given candidate digit, a node is "true" if any cell in the node contains the digit, and is false when no cell in the node contains the digit. For a link to exist between two nodes, all cells in each node must see all cells in the other.
When you say that the line of sight is reflected in some box, that is equivalent to saying that a link exists between two nodes lying in the box.
Code:  * * .2 * 7* * 3
* * .* * .6 2 .
2 3 .6 C 84 D .
++
7 6 .8 2 91 . .
* * .3 7 52 8 6
. 2 84 1 6. 7 9
++
. 7 95 . 2. 6 .
. 5 2A B .7 * .
8 . .7 B .* * 2 
Here is the formulation of one of the empty rectangle eliminations expressed in terms of a grouped AIC. I've labelled the nodes with letters A (r8c4) through D (r3c8). In this case, node B is the only grouped node. We see that node A is strongly linked to node B, which is weakly linked to node C, which is strongly linked to node D. Therefore r8c4 and r3c8 are strongly linked, so the candidate in question can be eliminated from any cell which sees both r8c4 and r3c8.
Just for exercise, a longer chain chain using boxes 1, 2, 3, and 8 (as your example does) is shown below:
Code:  F F .2 C 7G G 3
E E .D C .6 2 .
2 3 .6 C 84 H .
++
7 6 .8 2 91 . .
* * .3 7 52 8 6
. 2 84 1 6. 7 9
++
. 7 95 . 2. 6 .
. 5 2A B .7 * .
8 . .7 B .* * 2 
Here we have the chain
A = B  C = D  E = F  G = H
where "=" represents a strong link and "" represents a weak link. This chain likewise shows the existence of a strong link between r8c4 and r3c8. A point to remember is that a strong link can serve as a weak link in an alternating chain. In the above, the link between B and C is a strong link, but it serves our purpose to use it as a weak link in order to maintain the alternation of strong and weak links. Another comment is that there are sometimes options as to how you partition the cells within a box into two nodes. For example, it would have been acceptable to consider r8c45 as node A and r9c5 as node B. Just remember that for a cell to "see" a node, it must see all cells in the node. Note also that in this alternate grouping, the link between B and C would not be a strong link, but that's OK, of course.
These links may be of interest to you:
Alternating Inference Chains: http://www.sudoku.com/forums/viewtopic.php?t=3865
Grouped XCycles: http://www.sudoku.com/forums/viewtopic.php?p=17612
My post in this thread, and Myth Jellies' response (grouped AIC's in Nightmare solution):
http://sudocue.net/forum/viewtopic.php?t=118 

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istvan New kid on the Grid
Joined: 10 Nov 2006 Posts: 4 Location: Budapest, Hungary

Posted: Mon Nov 13, 2006 10:04 pm Post subject: 


Ruud, Ron, thanks guys. So if one finds such reflections chances are that there are some fishes close there.
(By the way, coloring the beam in both directions, I mean starting from one end of the strong link and marking all true candidates, then starting from the other end and marking true candidates the opposite direction, all candidates along the beam that are not marked can be eliminated as they could never be true. At the first example it would eliminate the same candidates as the swordfish; I don't know whether it would be always true. But this coloring needs more attention, not so easy to spot as the original reflection pattern in that example.)
I liked the AIC explanation very much, that is a very clean theoretical background for such patterns. Thanks again.
Istvan 

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