Comments - Talk
... by: ralph maier
is a restricted common candidate necessary for ALS towork??
... by: sunshine48
A simple explanation, I finally got it. Wonderful site.
... by: Douglas Boffey
There is a second way a set can be almost locked, namely, when N candidates appear in N + 1 cells within a unit. I call this a Hidden ALS (or HALS), as knocking out one of the cells reduces to a hidden pair/triple/quad/&c. Likewise, the ALS described on this page would be better called a NALS (Naked ALS).
... by: Jason
Ian,
If I understand correctly, adding G5 to the brown ALS does no good. You can only use this technique to make eliminations on numbers common to both the ALS.
In Example 1, adding G5 to the brown ALS picks up another candidate, the 9, but it will not lead to any more eliminations because the yellow ALS does not have a 9. If the yellow ALS happened to have a 9, then it could help.
Essentially, the relationships between the common elements of the two ALS means that the unrestricted common candidate (in Example 1, it's the 8) has to be the correct solution for a cell in at least one of the two ALS. So at least one of the yellow or brown cells that contain 8 as a candidate must actually be a 8 in the final answer.
It logically follows that 8 can be removed as a candidate in any cell that can see ALL of the 8s in the yellow and brown squares.
... by: Ian Saliba-Curtis
Hi Andrew! Example 1: Could you not also have added G5 to the brown ALS and made it an almost locked triplet over 4 candidates?
If yes - is there a reason why you didn't? If no - please could you explain why not?
Many thanks! Your site is fascinating. Kind regards, Ian
... by: Marc
Ok, I've found out the answers myself...
Answer 1 = No, it's not true. A bi-value cell ALS can share the same box with one or more cells of the other ALS
Answer 2= Yes, they can.
... by: Marc
I have two questions:
If one of the ALS is a bi-value cell, then the other ALS cannot have a cell in the same box of the bi-value cell, is that true?
If none of the ALS is a bi-value cell, and they both are in a row (or both in a column), can both these rows (or columns) appear in the same band (or stack)?
... by: rlhaben
Actually, sorry. I was wrong on this one. It does present a contradiction. I stared at this for 15 minutes trying to see it. Wrote the comment, then looked again and saw it immediately. Looking forward to getting this on the iPad. :)
... by: rlhaben
Andrew must be busy on the mobile solver since I have seen no answer. But I think Grandad is right. If you load the example, you will not even get to ALS as an example for those cells unless you manually remove the 2. However, even in that case, the proof doesn't seem to work. If you make F5 a 3, it does not present a contradiction in the two ALSs. It may present a contradiction elsewhere. But this type of example trips me up. I bought the book and I love it. But I often find that I can't prove a path I follow using its rules. I usually import the puzzle that I can't prove and watch what the solver does. Sometimes that helps, sometimes not.
... by: Grandad
I am almost certainly not the first to ask but - in the 2nd example, surely F1 should have possibles 1, 2 and 3 so isn't an ALS ?
... by: Terry
Hi Andrew,
Excellent site ;-) I'm just getting to grips with some of the more difficult strategies. I had a puzzle loaded in the solver and it came up with an ALS that I fail to understand :( The starting point for the puzzle is
..7...32.1..2.7..6.26.5.71..8...2....19.6.5...6.7......91.2..475..17.....78......
If you just keep clicking 'Take Step' until it comes to the first ALS, I can't see how 8 is restricted as 6 can also be seen by both ALSs.
What am I missing in my understanding of the ALS technique?
Many thanks and best wishes,
Terry
... by: Ana
In the last example the {1,2,6,7,8} in row G is not an ALS because it has 5 candidates in 3 cells
Andrew Stuart writes: The four cells highlighted in brown are the ALS, so thats 5 candidates in 4 cells. |
Post a Comment using Facebook...
|