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 strongly-linked) pair in r3c58. Alternatively, you can view the eliminations as coming from finned X-wings.
Code: Select all
* * .|2 * 7|* * 3
* * .|* * .|6 2 .
2 3 .|6 * 8|4 * .
-----+-----+-----
7 6 .|8 2 9|1 . .
* * .|3 7 5|2 8 6
. 2 8|4 1 6|. 7 9
-----+-----+-----
. 7 9|5 . 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: Select all
* * .|2 * 7|* * 3
* * .|* * .|6 2 .
2 3 .|6 C 8|4 D .
-----+-----+-----
7 6 .|8 2 9|1 . .
* * .|3 7 5|2 8 6
. 2 8|4 1 6|. 7 9
-----+-----+-----
. 7 9|5 . 2|. 6 .
. 5 2|A 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: Select all
F F .|2 C 7|G G 3
E E .|D C .|6 2 .
2 3 .|6 C 8|4 H .
-----+-----+-----
7 6 .|8 2 9|1 . .
* * .|3 7 5|2 8 6
. 2 8|4 1 6|. 7 9
-----+-----+-----
. 7 9|5 . 2|. 6 .
. 5 2|A 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 X-Cycles:
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