SudoCue Users

Hi

I am trying to understand the ALS-XZ pattern.
I have read the explanation a few times but it never really understood it. Now lately i made some eliminations that when checking it with Sudocue were mentioned there as Almost locked sets.
This made me think back how i made the eliminations and i think i have figured out how this technique works.

This is how i see it. Could you confirm if i see this correctly?

Locked set: N numbers in N squares.
Almost Locked Set: N+1 numbers in N squares.

You take 2 ALS's that do not use the same squares. When these ALS's both have one number that only appears in the same house within the ALS, you can use them for an elimination. Because they both have that one number in one house only one of them can contain this number. (As i understand this is then the shared digit)
Then you check what the implications are of that number not appearing in both ALS's (so turning them into a locked set). Then you compare these to see if there are any squares in which you can make the same elimination with either ALS (turned into a locked set).

I hope it is clear enough.

Is this way of thinking correct? This seems to me as a logical deduction using Almost locked sets. I was just curious if this is what is meant by ALS-XZ technique. Maybe there is more to it. If so can someone explain it to me.

greetings

Para

there is a dutch topic maybe you can write it there again in dutch because i don't know exactly what you mean here
Groetjes Marlie

I think what you’re saying is correct, Para.

Here's an example that might help - found by sudokuEd in a collaborative effort ( I helped a bit ) to crack Dec 18 X-Treme (which was a toughie)

This is a Sudoku-X puzzle so the cells in B are linked by the diagonal.
ALS(A=[r2c2,r3c3], B=[r3c7,r9c1], X=8, Z=9) => [r2c1]<>9

Group A and Group B share the candidate 8 - it cannot be in both at the same time therefore one must lose it. When this happens it forces the other shared candidate - 9 - into either or both of the groups therefore you can eliminate any 9 that sees both ie r2c1<>9

Para wrote:You take 2 ALS's that do not use the same squares.

It sounds like you've got ALS Para, especially when you describe this crucial bit about ALS.

I still find ALS eliminations very hard to find after a couple of months trying - since they can involved any number of cells. If you find a quick way to spot them, please share.

emm wrote:Here's an example .... to crack Dec 18 X-Treme (which was a toughie)

. Don't agree emm, Dec 18 was pretty easy. Dec 17 was the toughie

Cheerful after a wonderful holiday in NZ
Ed

ps. There is another (much easier) way to crack that spot on the Dec 17 puzzle - isn't there emm?

Yes there is Ed - and if we’d read the Solving Guide properly we’d have known it before - ALS logic is the same as XY / XYZ wing but the multivalue ones are still really hard to spot.

Also you’re right, it was the 17th that was tough - the 18th was a doddle ... as were the 19th, 20th, 21st… perhaps Ruud is going easy on us for Christmas