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 Post subject: Killer ALSxz
PostPosted: Tue Apr 13, 2010 9:33 am 
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This is a new thread about what Mike has nicely dubbed the Killer ALSxz technique. I will just repost my examples given in another thread. Mike will repost his definition and explanation following.

Example 1
....... this example has.. one cell that is a peer between two remote cages. This forces an interaction.

Image

I'll quote Afmob's PM to explain this one.
Quote:
14. "45" on n9: 3 innies r7c9 + r89c7 = 17 and must have 9 for n9 = [2]{69}/[5]{39} = [5/6..]
14a. 14(2)r7c56 = {59/68} = [5/6..] -> Both cages can't have 5 (at the same time) since cells containing 5 see each other (r7c569) -> At least one cage must have 6 -> CPE: no 6 in r7c8

Note the bold part. This makes the move work. If it were only a Killer pair move then it would be cells containing 5 and 6 see each other, so that one cage must have 5 and the other must have 6. This is not the case here because both cages can have 6 (and indeed, both will have 6!).


Example 2
This example is more complex because it involves one cage and one outies "cage" but again, they have one cell that is a common peer.

Image

12. Outies n9 r9c6 + r6c9 = {46}/[19] = [6/[1]..] ie, they have 6 or 1
12a. 8(2) at r6c6 = [71/62] = [6]/[1] ie, they have 6 or 1
12b. Both cages can't have 6 (at the same time) since one cell containing 6 sees all cells of the other, ie r6c6 sees both r6c9 & r9c6 -> At least one cage must have 1 (and indeed, both will have 1!) -> CPE: no 1 in r89c7

Cheers
Ed


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 Post subject: Re: Killer ALSxz
PostPosted: Tue Apr 13, 2010 12:52 pm 
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Quote:
This is a new thread about what Mike has nicely dubbed the Killer ALSxz technique.

Yes, as I said on the other thread (here), we should re-use the existing "ALS-XZ" nomenclature (or "ALSxz", as Ed writes it), because it's the Killer equivalent of the established ALS-XZ technique in regular Sudoku.

The technique relies on finding two groups of cells, both having the property that they must contain at least one of the digits {XZ}, where all cells containing the candidate X in both groups see each other. As such, only (at most) one of the two groups can contain X. The other one (whichever it turns out to be) must therefore contain a Z. In other words, the two groups must contain at least one Z between them. We can therefore remove the candidate Z from any cell that sees all of the cells containing the candidate Z in both groups.

Each group involved can be either (i) a cage, or (ii) an almost locked set (ALS). I suggest using the term "Killer ALS-XZ" for the cases where at least one of the groups is a cage, dropping the word "Killer" from the name only if both groups are ALSes (corresponding to the standard ALS-XZ scenario in vanilla Sudoku).

Note that one of the two groups can be a single bivalue cell containing the candidates {XZ}, since it also conforms to the generic definition of an ALS as being any group of N cells that contains (N + 1) candidates. Indeed, from a logical perspective, a cage that must have at least one of {XZ} behaves as if it were such a bivalue cell. However, due to the different geometrical considerations, a Killer ALS-XZ move with two cages is possible (as shown in Ed's examples), whereas an ALS-XZ move with two bivalue cells isn't (it always degenerates into a naked pair).

In the case of cages, the condition "must contain at least one of the digits {XZ}" is a convenient one, because this information often drops out of the combination analysis. For example a 19(3) cage with the combinations {289/379/469/478/568} must contain at least one of {89}.

In the case of ALSes, the condition "must contain at least one of the digits {XZ}" derives from the fact that an ALS (being only 1 candidate away from being a locked set) can only afford to "lose" one candidate digit. Therefore, if an ALS contains (possibly in addition to others) the candidates {XZ}, we know in advance that the cells involved must contain at least one of {XZ} in the final solution.

Incidentally, the digit X, where all cells containing this candidate in both groups see each other, is typically referred to in the literature (and in my walkthroughs :) ) as the restricted common. This is because it can only be present in one of the two groups. Note that if Z is also a restricted common, a Killer ALS-XZ "degenerates" into a killer pair on {XZ}.

Ed has already supplied a couple of nice examples where both groups are cages, so here's an example (taken from an unpublished puzzle) of a Killer ALS-XZ involving a cage and an ALS, which should make the connection to the established ALS-XZ technique more apparent:

Image

Here, the 17(4) cage at R3C5+R4C456 has the possible combinations {1259/1349/2357}. In other words, it must have at least one of {29}. Likewise, the ALS at R9C457 = {2679} must have at least one of {29}. Furthermore, all cells containing the candidate 9 in both of these groups (i.e., R49C4) see each other. Since only one of the groups can (because of this) contain the digit 9, at least one of the two groups must contain a 2. We can therefore remove the candidate 2 from all common peers of all cells containing the candidate 2 in both cages (R4C6+R9C5). In this example, this results in the 2 being eliminated from each of the yellow cells, resulting in a naked single at R9C6 = 3.

Describing such moves in walkthroughs is a matter of style. In my own walkthroughs, I refer to Killer ALS-XZ moves along the following lines (using the above example):

Quote:
1. 17(4) at R3C5 = {1259/1349/2357} = {(2/9)..}, with 9 only available in R4C4
1a. R9C457 = {2679} (ALS) = {(2/9)..}, with 9 only available in R9C4
1b. -> 17(4) at R3C5 and R9C457 form a Killer ALS-XZ on {29} with 9 as restricted common
1c. -> 2 locked in R4C6+R9C5
1d. -> no 2 in R5C5+R9C6 (common peers)

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Cheers,
Mike


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