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# Aligned Pair Exclusion

This is an interesting strategy, known by the short-hand as APE and sometimes called Subset Exclusion. It can overlap with Y-Wings, XYZ-Wings and WXYZ-Wings but uses very different logic. The overlap is not strict so they are worth looking out for in a tough situation.

There is always a base pair of cells (which now show up as grey cell on the solver). At least one elimination will occur in one of those two cells. The solver will also show a variety of colored cells which are the elements used to make an elimination. I used to distinguish between APE type 1 which only used bi-value cells and type 2 which used 2-cell Almost Locked Sets (ALSs). But the solver will now find a larger variety including 3-cell ALSs and since these merely extend the same logic the solver will return the first of any it finds. A better Type 1 and Type 2 distinction is between the base pair of cells which can be a locked pair (ie can see each other) or not (can't see each other). The logic is subtly different but I'll come to this in the following examples.

## Aligned Pair Exclusion - Type 1

The Aligned Pair Exclusion can be succinctly stated: Any two cells that can see each other CANNOT duplicate the contents of any Almost Locked Set they both entirely see and share candidates with.

Remember - a bi-value cell (with two candidates) is the simplest Almost Locked Set since it is a set of size '1' with size+1 (ie two) candidates.

Let's consider the simplest possible example - two bi-value cell attacking the pair. I have also shown the Y-Wing in the diagram so we can see there is a simpler way to do the same job - but only in some cases.

We consider ALL the possible pairs of numbers that will fit in [G2/G3]. These are for G2 and G3:

2 and 2 (impossible)
2 and 5
2 and 8
4 and 2
4 and 5
4 and 8

Apart from the first being impossible (2 and 2) since G2 and G3 can see each other, we have problems with some of the other combinations. What if 2 and 8 were tried as the solutions? Well, that would duplicate and therefore empty G9. Also 4 and 8 would empty H1.
We are left with a set of combinations that looks like this:

2 and 2 (impossible)
2 and 5
2 and 8 (impossible)
4 and 2
4 and 5
4 and 8 (impossible)

Notice that we now have no 8 left in any pairing? Therefore we can remove 8 from our base pair. Voilà

## Example 2

The next example has tri-values spread over two cells as part of the attack. The way we can use double cells is by saying that any two cells with only abc excludes combinations ab, ac and bc from the base pair under consideration. This neat trick greatly extends the usefulness of APE which would otherwise be a just poor man's Y-Wing.

The 2-cell ALS in [A1,B3] contains {1/3/7} so pairs that would cripple the solution for that ALS are {1,3}, {1,7} and {3,7}.

Let's consider all the possible pairs of numbers in our base pair [C2/C3]. These are:

1 and 3 - excluded by A1 + B3
1 and 4 - excluded by C5
1 and 9
3 and 3
3 and 4
3 and 9
8 and 3 - excluded by C9
8 and 4
8 and 9

Now, we have to be a tiny bit careful here. 3 has definitely been excluded as a possible solution in C3 but look down the list and 3 + 4 is still OK and 3 + 9 is OK. So we can't remove 3 from C2 just yet.

Credits: Myth Jellies came up with the insight for abc = ab/ac/bc

Note: There could be more than two, sometimes three or four ALSs of several sizes in an APE attack. I've considered examples with two for simplicity's sake

## Example 3

Further into the same puzzle we come across a 3-cell ALS plus a bi-value in H3 attacking A3/B3. The ALS in [A1,C1,C3] contains the four numbers {1,4,7,9} which the solver thinks of as a quadruple combination. The combinations of 'abcd' are ab, ac, ad, bc, bd and cd. Back to the base pair: We can list the combinations for A3/B3 as

4 and 3
4 and 7 - excluded by [A1,C1,C3]
5 and 3
5 and 7 - excluded by H3
7 and 3
7 and 7 (impossible)

The tricky one with the 3-cell ALS is not the fact that the base pair will empty it (it can't since it is two cells and the ALS is 3 cells). It's the fact that a solution of 4 in A3 and 7 in B3 would mean there'd be only two candidates left to fill three cells. Thats enough to rule out the combination.

## Aligned Pair Exclusion - Type 2

Aligned Pair Exclusion can also work even if the pair is not aligned. Sounds like a joke, but it's too late now to rename this strategy :) Perhaps 'Subset Exclusion' was a better idea. There is a subtle logical different but I have found many examples and it boosts the usefulness of this strategy.
I'm very grateful to Joseph Aleardi for putting me on the scent of this elegant logic.

The simplest type of APE2 using just two bi-value cells duplicates the Y-Wing, but I include an example to illustrate how APE2 works.

The diagram here shows first the Y-Wing based on A1 - A4 (the pivot) - B6. It's quite easy to see that 8 must occur in either A1 or B6, thus removing it from B1 and B2.

But let's follow the APE logic with the non-aligned pair A4 and B1. (Note: We could also choose A1 and B2 and eliminate the 8 there also). A4 pairs with B1 using these combinations:

1 and 1 - POSSIBLE!
1 and 6
1 and 8 - excluded by A1
9 and 1
9 and 6
9 and 8 - excluded by B6

The only difference between APE 1 and APE 2 is that with non-aligned pairs the same candidate *could* be a solution in both cells. So 1 and 1 is definitely on the cards. Not that it is critical in this case. The other exclusions mean we can't have an 8 in B1, just as we thought.

Here is a more complex APE that does not have a Wing alternative. We have two bi-value cells and one two cell ALS attacking B1 and C7. Let's write out the combinations between those cells:

1 and 4 - excluded by B9
1 and 5 - excluded by B8
1 and 7 - excluded by [C1 + C3]
2 and 4
2 and 5
2 and 7
7 and 4
7 and 5
7 and 7 - Permitted!

Clearly 1 is removed from B1. The exact same formation also removes 1 from B2.

To conclude, a non-aligned pair using a bi-value cell and a 3-cell ALS. I'll leave it to the reader to work out why 6 can be removed from D1.

There is a second very nice APE later in the solving sequence using a 2-cell ALS and a 3-cell ALS. You can load the puzzle from the links under the diagram.

## An Eight-Cell Aligned Pair

I love to end these articles with a Sudoku from Klaus Brenner. He has made finding interesting and beautiful examples an art form and in the case of Aligned Pairs has found what we thought was impossible. An eight-cell Aligned Pair elimination! We had found some five-cell examples and wondered if there could be a six-cell or even a seven-cell. This is the first and only eight-cell formation known. Fortunately the solver can handle this many. Each cell is necessary to produce all the pairs used to cross reference with the target cells in D5 and E5 - the solver ignores any other ALSs that are not used. And very pretty it is.

How many difficult puzzles did Klaus have to check to find this? Astonishingly, around 21 million!
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## ... by: Pulsar

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.

## ... by: Robert

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 :)

## ... by: TAU

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.

## ... by: rRcCV

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.

## Monday 10-Mar-2014

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.
Andrew Stuart writes:

Correct

## Monday 25-Jun-2012

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).

## Friday 22-Jun-2012

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)

## ... by:

I want this software. how I can have it?
Andrew Stuart writes:

## ... by: Phil Gooda

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?

## ... by: Patrick Barnaby

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.

## ... by: Matt Lala

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.

## ... by: Bernard Gervais

align pair type 2
I found more than one numerical solution for this example.
Best regards.
BG

## ... by: Bernard Gervais

Excellent site, thank you.
for example 1, I use the unicity concept which pinpoints A4 = 1.
Best regards.
Article created on 11-April-2008. Views: 198248