From Sudopedia, the free Sudoku reference guide
Talk:Aligned Pair Exclusion
From Sudopedia
I hope this doesn't come across as hostile, but I have to say I'm a bit baffled by what I see here. I do remember seeing the APE term when I was trying to acquire new Sudoku solving skills, and, as noted in the article, I considered it too tedious and went on to other techniques. Now that I feel comfortable working with Almost Locked Sets, I don't see any need for this. APE may have had its place at the time it was discovered, but we shouldn't continue using old tools when there are simpler, more effective, and more general tools which accomplish the same and much more.
As the article states, many APE positions can be considered as XY wing or XYZ wing patterns. Of course, these are simple examples of ALS XZ rule patterns, and all the examples I've seen of APE can also be considered as ALS XZ rule patterns, including this one. Here we can use:
- r4c3 as ALS 1, 1 cell with 2 candidates, {1,9}
- r126c1 as ALS 2, 3 cells with 4 candidates, {1,2,5,9}
- X (restricted common digit) = 1
- Z (eliminated digit) = 9
This gives the elimination of digit 9 in r5c1, as shown in the article, as well as the 9 in r2c3, not shown. Of course, in this case the latter elimination follows quickly from the first, but in more general positions there could be other 9's in box 4 so that r4c3 could not be immediately solved as 9. For example, suppose that r4c1 was not a solved cell, but contained 9 as a candidate. The ALS XZ rule would give the elimination of this candidate as well, whereas APE would again require the tedious listing of pairs of values from r4c1 and r6c1.
This example is also rather convenient in that r5c1 is bivalued. Suppose in a more general position that it had, say, seven candidates, with 9 among them. Can considering all the 21 possible pairs from r5c1 and r6c1, and then observing that all pairs with a "9" from r5c1 lead to a contradiction, be a good way to approach this? Of course, with larger ALS's r6c1 could have more candidates, too.
I can't say I've seen that many examples of APE, but all I have seen can be viewed as containing an ALS XZ rule pattern, and any generalizations or subcategories seem to differ only in the size of the ALS's involved. If somebody can provide an example of APE, or aligned triple exclusion, etc. which doesn't lend itself to an ALS XZ rule argument, or perhaps a more general ALS argument, I'll gladly offer an apology. In that case, we should attempt to characterize the position to explain why it doesn't lend itself to an ALS argument -- then we'll have a real use for it.
I admit that some of the material here might be a bit dated -- most of the material was entered several months ago, and then Ruud and I both seem to be busy and only a few new articles get created since then. Would be better if more people actually improved the articles here. =)
Back to your comment on APE and ALS-XZ:
----------------------------------------------------------- | 239 8 23469 |#46 5 1 | 7 36 39 | | 39 1 369 |#68 2 7 | 5 4 389 | | 5 7 46 | 9 3 468 | 168 168 2 | |-------------------+-------------------+-------------------| | 4 2 5 | 7 689 689 | 1389 138 138 | | 7 39 8 | 134 149 2 | 49 5 6 | | 6 39 1 | 5 489 3489 | 489 2 7 | |-------------------+-------------------+-------------------| | 239 6 239 |*1348 14789 5 | 138 1378 1348 | | 1 5 39 |*348-6 46789 34689 | 2 3678 348 | | 8 4 7 | 2 #16 #36 | 136 9 5 | -----------------------------------------------------------
I have long thought that APE (in its unextended form, see the Scanraid article for the extended form) can always be replicated with ALS-XZ. Until I ran into the above example, where the aligned pair is r78c4 and the conclusion is r8c4<>6. I might be wrong, but can you find an ALS-XZ that leads to the same elimination? Of course, ALS-XZ can be applied to eliminate 1 from r7c5. =P
If someone can prove that APE/ATE can be replicated using some other ALS technique, then we can simply say so in the introduction of this page and leave the remainder of the page as it is as some kind of historical reference.
--unkx80 12:18, 30 April 2007 (CEST)
A further thought on APE versus ALS-XZ:
.-------.-------.-------. | A B | X X X | X X X | | | | | | Y | | | :-------+-------+-------' | | | | | | | | | | | | :-------+-------+-------' | | | | | | | | | | | | '-------'-------'-------'
Suppose we enumerate the values in the cells A and B and due to the bivalued cells marked X and Y, some candidate is eliminated from A. Here, the cells X belong to the same row as A and B while the cells Y belong to the same box as A and B. If there is only one such cell Y, then definitely we can replicate this APE with ALS-XZ, where the two ALSs are (a) Y itself, and (b) B plus all the X cells. However, if there are two or more Y cells, then it might not be possible to find the required ALSs from the cells A, B, Xs and Ys that makes the same elimination.
--unkx80 12:44, 30 April 2007 (CEST)
Thanks for your response. I believe you're correct -- there's no ALS XZ rule elimination, but an AALS argument can be used, expressed by this Alternating Inference Chain (AIC):
- (648=(1or3))r127c4 - (13=6)r9c56 => r8c4 <> 6
We have an AALS in r127c4 which is doubly linked by the restricted common digits 1 and 3 to the ALS in r9c56. The extra restricted common digit compensates for the added degree of freedom in the AALS.
This points me to the approach I should have taken, but I'll need to give it some more thought on more general applicability. ALS XZ rule eliminations can always be expressed in terms of subset counting arguments, and this elimination can be also. Consider the subset of cells r127c4, r9c56. In total these five cells have five distinct candidate digits -- 13468. Digits 1 and 3 can each be placed at most once in the subset, since all of those candidates are in box 8, and digits 4 and 8 can each be placed at most once, since all of those candidates are in column 4. Without using digit 6, at most four cells could be filled, so digit 6 must appear at least once in the subset. Cell r8c4 sees all cells of the subset, so it must see a "6" somewhere and cannot therefore contain 6.
My frustration has been due to the poor examples I've seen, so thanks for this good one. I had checked the scanraid site, and found the first example to be an ordinary XY wing and the third to be an ordinary naked triple. The second example (extended APE) is non-trivial, but there's no difficulty expressing it as an ALS XZ rule position. It's an XYZ wing in which one of the normal bivalued wing cells has been replaced by a two cell ALS containing the two normal digits for that wing cell, plus another, of course. There's no trouble with the ALS XZ rule if either wing cell is replaced with a larger ALS containing the normal two values for that wing cell (in the simple XYZ wing configuration), as long as the restricted common digit is still preserved.
The general APE principle, if we're going to use it, should be something like this (edited):
If we agree that two sets of cells "see" each other if they are disjoint, and all cells of the two sets lie in some common house, then any set of cells which sees some Almost Locked Set can contain at most one of the candidate digits of that ALS.
The particular makeup of the ALS, in terms of the mix of bivalue, trivalue, ... cells does not affect the ALS XZ rule, nor the associated subset counting argument.
In the original post on APE, I had trouble understanding the ground rules for solving "unsolvable" puzzles. Both examples there could be handled with XYZ wings, and one could be advanced with an (independent) XY wing, but these evidently were not allowable techniques while this new (at that time) APE was (?). XYZ wings were known by that time, but perhaps not yet popular.
First of all, I'm sorry about giving no signature to my previous posts. I wasn't trying to remain anonymous -- I just haven't figured out everything yet.
I feel a bit silly, as shortly after I posted the above, I realized this was what I would consider an Almost Locked Candidates (ALC) position. The Sudopedia page on ALC is another one which needs some work, as Unkx80 has commented on the discussion tab for that topic. In this case, the ALC technique produces a much better yield than APE, or my earlier subset counting argument.
-------------------------------------------------------------- | 239 8 23469 | 46 5 1 | 7 36 39 | | 39 1 369 | 68 2 7 | 5 4 389 | | 5 7 46 | 9 3 468 | 168 168 2 | |-------------------+----------------------+-------------------| | 4 2 5 | 7 689 689 | 1389 138 138 | | 7 39 8 |*13-4 149 2 | 49 5 6 | | 6 39 1 | 5 489 3489 | 489 2 7 | |-------------------+----------------------+-------------------| | 239 6 239 |*1348 4789-1 5 | 138 1378 1348 | | 1 5 39 |*348-6 4789-6 489-36 | 2 3678 348 | | 8 4 7 | 2 *16 *36 | 136 9 5 | --------------------------------------------------------------
The ALC argument goes something like this:
In column 4, digits 1 and 3 are locked into cells r578c4. Both of these digits cannot fit into cell r5c4, so at least one of them must lie in the column 4/box 8 intersection. Whichever one does will combine with cells r9c56 to form a "136" triple which gives the eliminations shown for digits 1, 3, and 6 in box 8.
On the other hand, since r78c4 is a "set of cells which sees the ALS in r9c56," it can contain at most one of the candidate digits of that ALS, so in particular, at most one of the digits 1 and 3. Therefore, r5c4 must contain one of digits 1,3 and thus cannot contain 4.
I'm not yet sure that every APE or ATE example can be better handled with some less tedious and more productive technique. However, for myself at least, I would consider APE, ATE as "techniques of last resort." --Rmoore 19:37, 3 May 2007 (CEST)
