### ... by: ad.joe

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).

### ... by: ad.joe

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: 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.