It looks like all the given examples can also be solved using two Almost Locked Sets with restricted commons. Consider the most difficult example, the Eight-Cell Aligned Pair:

We can identify a 2-cell ALS {B5 C5} and a 4-cell ALS {D5 D6 E6 F4}. These two sets have a pair of restricted commons: 1 and 9. It is impossible for both sets to contain both digits, because within these sets they can only appear in the cells B5, C5, D5 and F4. These four cells cannot contain two 1's and two 9's, because the first three cells all see each other. Therefore, at least one ALS has to lose either a 1 or a 9 to become a locked set.

Now 2 becomes the 'pincer' digit: cell E5 can see all the 2's in both ALS's and can therefore not contain a 2; if it did, both ALS's would lose their 2's, but we already know that at least one ALS must lose a 1 or a 9, and an ALS can only lose 1 digit.

You can apply the same strategy with the 2-cell ALS {D6 E6} and the 4-cell ALS {B5 C5 D5 H5}. In this case, 6 and 7 are the restricted commons.

Two Almost Locked Sets with restricted commons appear to cover a wide variety of cases. I'd love to see if they encompass all forms of aligned pair exclusion.

Interesting strategy. I do have some thoughts though.

All of the examples, with the exception of the last, are cases of an AIC with almost locked sets. My own solver, which implements this without a size limitation on the ALS (the solver at this website seems to consider only ALSs with two cells and three values) does not solve the Klaus Brener 8-cell example; however, my cell-forcing implementation does.

I think there is another way to view this sort of strategy. There are a number of ALSs, and also some AALSs shown. The logic of an ALS in an AIC (described on another page is this) - if a chain turns a value in an ALS "off", by virtue of being weakly linked to something somewhere that has already been turned "on", then the other values in the ALS are all turned "on" - that is, the values in an ALS are strongly linked (but not weakly linked) to each other. Or, put another way, once a value is turned off, the ALS becomes a locked set, and the values in the locked set can be eliminated from other cells in the unit containing the ALS that was promoted to a locked set. The chain can then continue, and possibly allow us to draw an inference.

But this idea can be extended. Suppose you have an AALS - then two values need to be turned "off" in order to turn it into a locked set. This will happen quite a bit in a forcing net (the only significant strategy I have implemented in my own solver, that is not described in detail at this web site). However, it could even happen in an AIC - the chain could turn "off" some candidate in the AALS, then wander around for a while, and come back and turn off another candidate in the same AALS. Two is all that is needed - the AALS has now become a locked set, and the inference can continue.

I have been calling this a "dynamic locked set" - it isn't a locked set, but conditional on some assumption (an AIC or forcing net becomes simply by assuming some candidate is on or off), once all the implications of that assumption are considered, the "dynamic locked set", which might have had one, two, or even more values than a locked set would, magically becomes a locked set. And we can then turn "off" the remaining values in the dynamic locked set if they are found in other cells in the unit, and possibly continuing following the chain/net until it allows us to make some inference.

I have similarly implemented a forcing net with what I call "dynamic groups", "dynamic fish", and "dynamic finned fish". I tried "dynamic Y-wings" and "dynamic XYZ-wings", but in my small database of 245 advanced puzzles, but this resulted in exactly zero additional inference that the forcing net could not already make. In fact, I think among the strategies described at this web site in the "Basic", "Bent Set", "Strategies", and "Forcing Chains" sections, the only ones which are possibly not special cases of "forcing net with dynamic groups/fish/finned fish" are AIC with unique rectangles, and SK loops. Even the "exotic" strategies include some special cases - I'm not sure about pattern overly, but other than that and Franken Fish, all "exotic" strategies listed up to APE are special cases.

So I'd say you could view this one is a particular case of "forcing net with dynamic locked sets". That is meant in no way to denigrate this quite elegant strategy; I just think it can be generalised.

Franken fish is my next project :)

if I hit in APE Example 3 , it will be changed in the solver, different from what I can see in this picture.
Best regards
Giuseppe Rizza

I've found a very unusual APE example - try this one

The only other thing that happens here is Hidden Singles, and on the step with the APE 4 different ALSs are used to eliminate a total of five candidates from the two grey cells, which is definitely really high.

I need some help, because I thought I understood WXYZ-wing, but in example 2, JUST before the APE would kick in, wouldn't it be a valid WXYZ-wing with A3 as hinge, together with A1/A4/H3? Locked Set of 1/4/5/7. It would clear 7 from B3, therefore solving it as 3.

* If A3 is 4, A4 solves to 1, which leads A1 to be solved as 7, which removes the 7 of B3
* If A3 is 5, H3 is solved as 7, which removes the 7 of B3
* If A3 is 7, directly removes the 7

The 7 in question is removed anyway by a second APE after the APE and Simple Colouring, but is removed because the combination 5/7 in A3/B3 would clear H3... which is part of my possible WXYZ-wing, together with A3.

Is there an assumption of WXYZ-wings that I have somehow missed?

I don't know whether your solver implements it, but the article on APE Type 2 does not mention that the combination of the target value with the same value in a non-aligned pair cell CAN be excluded, not by an ALS, but by a Locked Set, IF the target and pair cells, between them, can see all occurrences of the common value in the Locked Set.

Like your site very much.
But I would appreciate to get some help in understanding the last "Eight-Cell Aligned Pair" example.
As far as I can see eliminating candidate 2 from E5 just requires four of the eight cells:
B5, C5, D6 and E6.
And completing the exercise of checking all possible combinations in D5E5 against the impact of all the ALS highlighted in the diagram unveals a lot of other combinations as being impossible, but they are not enough to eliminate any candiates from either D5 or E5, since I#m left with the remaining combinations as follows:
1 7
2 4
6 9
7 9
9 7
So what's the benefit of taking EF4 and HJ5 into consideration?
I just don't see what makes the statement true that "Each cell is necessary to produce all the pairs used to cross reference..."
Thanks for any explanation.

I think it is important to note that any ALS seen by both cells doesn't have to be entirely seen by both cells.

For example if we were excluding "3 and 5" then the first cell only has to see the cells in the ALS that can be 3 and the second cell only has to see the cells in the ALS that can be 5.

We are great fan of your site and tools, which we use for training, and I for some leisure. I am not an expert on sudoku, but I am something of an expert on training people to do complicated procedures. In my work, is we fail at a the training, later, either something/someone blows up (quite literally, I'm afraid), or a 1M$+ piece of equipment is taken offline or ruined.

The following comments are offered, then, as a more general comment (one example), to challenge your process of explaining how one begins to apply advanced "strategies". It is communicated because those students and I using your clearly thoughtful and effort-intensive explanations find it possible, even so, to apply them, in largest part, only if the explained strategy is already understood beforehand/separately from the explanation you provide. It is offered in the spirit of good organizational practice—that it is better we be talkers (providing critical feedback, even if negative) than walkers (departing the organization without comment, when needs are not met).

First, note, in your pedagogy of advanced strategies, you *do* explain superbly how the pattern you choose to act upon leads to elimination of possible numbers in cells, based on the new method being applied—the examples are all, generally, excellently explained.

But, there is little or no explanation on how one engaging a puzzle at an advanced state of solving (that shows all possibles, and that is taken as far as preceding more basic techniques allow) can discern either that no other advanced strategy is better and actionable, or that the currently presented strategy, here APE, is present and actionable, and if several starting points are possible, how the specific one to act upon is chosen.

Hence, what we seek (elsewhere at present) are explanatory statements for various strategies, that indicate, for given points on the paths to a puzzle's solution, what patterns are to be looked for, among which frequencies of remaining numbers/number groups, to indicate the best advanced strategy to attempt to apply.

In APE Example 1, perhaps what is sought is as statement like, "After scanning the puzzle for… [patterns] and seeing none further—note that the apparent starting points for strategies P and Q in cells A and B are dead ends—it is then may be time to apply an Aligned Pair Exclusion. To see if this strategy is applicable, units with between 4 and 5 unsolved cells are examined for presence of 2 or more bi-value cells that can *see* the rest of the cells, and that *share candidates with them*. Of the possible starting points in Example 1, we choose row G and box GHJ123 because… and we focus on cells G2 and G3 because… " [Note, no mention at all, yet, of Almost Locked Sets of greater complexity than bi-value.]

A related point here is that a very broad and general statement of the strategy is likely most suitable at the end, where it can be readily understood, e.g., here, where the extension from bi-value to all other Almost Locked Sets can be explained without undue complexity.

Remarkably, we know that you have already thought through this logical, truly strategic decision-making—it is what your programming, in the solver, clearly already does, walking stepwise, in a particular order, through the strategies, deciding which to apply or dismiss, based on looked-for patterns. It is this very process of logic, of what to apply when in the course of a puzzle, and where (very specifically) to begin applying it within a puzzle matrix, at a given point of being solved, that is missing in the current "strategy" explanations.

Finally, we are aware that explaining "strategies" is not necessarily the same as explaining the process of developing an over-arching strategy of applying them, in orderly fashion to given situations. In this regard, I would say you have laid out an excellent list of *tactics*, and that what is yet missing is an adequate discussion of strategy—the way in which "battlefield" conditions are accessed in each situation, to decide which particular tactic best applies.

Hope this might be go some help. We will continue exploring and learning tactics, separately, and coming here to review your well-explained examples, and to sort on our own the best over-arching strategies to apply. Best wishes for this continuing, clearly dedicated and effort-intesive endeavour. LeProf

General structure of the APE
------------------------------------
Maybe it's not easy to put this straight, but in fact there are eliminations in 2 boxes:

g k p | _ _ _ | Y Z _
_ _ _ | _ _ _ | _ _ _
_ _ X | _ _ _ | a b c

The "wings" are X=2#, Y=2*, Z=2+

Then XYZ eliminates 2 on the aligned cells abcgkp ...

(provided that these have "not much more than #*+ on it", so you have to "cross-check")

- Doesn't the APE-structure look like an xy-wing (y-wing)? Only the function does work backwards and there is no pivot necessary!!
So you could also call it a BACKWING or a BACKSWING!

- And of course, in an X-Sudoku the 2 aligned fields need not be in the same box
(as long as one of the two is on a diagonal).

Let's build a general structure for the APE
_ _ _ | g k p | y z _
_ _ _ | _ _ _ | _ _ _
_ _ _ | x _ _ | a b c
the "wings" xyz (x=2#, y=2*, z=2+) eliminate 2 on the aligned cells abcgkp
(of course also in box 2, not only in box 3)

- While seeing this structure we can say, the form looks the same as an xy-wing (y-wing) or an xyz-wing, but the function works backward.
So you could call it also a BACKWING or a BACKSWING

- And of course, in an X-Sudoku the aligned fields need not be in ONE box (as soon as one of the two is on a diagonal)

How to tease out the APE:
-----------------------------------
First, it's not our job to solve one of the examples in another way. But it's nice that all is so understandable.
(Matt, I'd add three words: duplicate the contents of any "ONE of the" two-candidate cells they both see. )
So, as we look for the third cell of an xy-wing (having seen two "similar" ones at first), what can we do here:
Let's solve it by cross-checking: We assume that the most FREQUENT number of the bivalue cells can be eliminated in the two cells in question.
We assume this number being correct there and quickly end in error, quicker than in looking up all combinations.
(Who'd think that we would not score more than 90 percent, he/she can feel free to try with the other numbers)

I want this software. how I can have it?

I actually managed to understand all this, except for one thing......why bother, there is a far simpler soultion within that box? Let me demonstrate by considering ONLY that box and naming the cells as follows:

A B C
D E F
G H I ..... which means that my C is the X mentioned, and my F is the Y mentioned.

Cells B, C and G contain 3 numbers and only 3 numbers between them. Therefore those 3 numbers cannot appear in the other cells. Which immediately gives you the paired 5/9 in F and I, which means that E has to be 4. Why bother doing all the APE stuff?

An XY-Chain removes 7 from r8c8. A Bilocation Graph reveals a 2 at r7c2 and then you'll have a naked 7 and then a naked 8 and then quite a few hidden singles on the first and second rows. So this is easier than APE except the Bilocational graph techinque.

I love the site but some of the explanations need help. Or I guess I do. I think this is one of the less clear ones.

The rule you have says:
Any two cells aligned on a row or column within the same box CANNOT duplicate the contents of any two-candidate cell they both see.

If I take that literally in the first example... X in the first example can see the bivalues 25 and 37. It cannot duplicate the contents of those cells. Therefore... it can't be 2,5,3 or 7? Obviously that's not what was meant.

Or is it... if the two aligned cells see some bivalue cells, and they both mutually share a certain candidate [that's part of those bivalue cells] then that shared candidate can go? But no, that's not it either.

If either if the cells see two bivalue cells, and those two bivalue cells both share a candidate with the cell under consideration... that shared one goes?

Does one actually have to enumerate the possibilities? This seems like something that can be spotted with a glance, if only it can be made clear the exact conditions needed.

Excellent site, thank you.
for example 1, I use the unicity concept which pinpoints A4 = 1.
Best regards.