ronk wrote:Sorry, I stopped reading too soon, Ron

De nada, i enjoyed this coincidence

17 posts
• Page **2** of **2** • 1, **2**

Thanks to all, so i have 3 solutions using an ALS now, which i think, all can be spotted easier than the 3 forcing chains (together) that susser used. I wondered, if i should change my solution method, which is primararily based on searching for forcing chains after basic techniques. Mabe i should look for ALS first? But after going through some old puzzles, i saw that i would have found easy ALS's as part of forcing chains using pairs. The advantage of ALS, that in many cases more than one elimination can be made, is also not that big, because when once i have a forcing chain with ALS, it would be easy to see the other digits i can kill with the same (main part of the) chain.

E.g. an xz-ALS elimination could be easily notated in a nice loop (though i dont know, if this is a correct notation, because =x=A in this case means "x must be in one of the cells of A"):

X(-z-A=x=A)-z-B=x=B

That means, there is a contradiction, if A and B do not have a common cell containg x.

So, what i am going to do in the next time, is to concentrate more on "suspicious" almost locked sets (with more than 3 cells), that i could use (similar to what i saw Carcul doing with AUR's).

E.g. an xz-ALS elimination could be easily notated in a nice loop (though i dont know, if this is a correct notation, because =x=A in this case means "x must be in one of the cells of A"):

X(-z-A=x=A)-z-B=x=B

That means, there is a contradiction, if A and B do not have a common cell containg x.

So, what i am going to do in the next time, is to concentrate more on "suspicious" almost locked sets (with more than 3 cells), that i could use (similar to what i saw Carcul doing with AUR's).

- ravel
**Posts:**998**Joined:**21 February 2006

17 posts
• Page **2** of **2** • 1, **2**