Almost Locked Sets

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David Bryant
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Almost Locked Sets

Post by David Bryant »

I'm still trying to understand the "XZ rule" thoroughly.
Solving Guide wrote:An almost locked set consists of N cells with exactly N+1 candidates.
So I was working the "nightmare" puzzle for Monday, 2 October, 2006. Some fairly basic stuff got me to this point, where I applied a "Double Implication Chain".

Code: Select all

*-----------------------------------------------------------*
| 46*   7     14B   | 9     25    3     | 145A  8     2456  |
| 469A  39    5     | 278   1     78    | 34    46A   27    |
| 8     2     13    | 4     57    6     | 9     15    1357  |
|-------------------+-------------------+-------------------|
| 57    14    2     | 36    36    1457  | 145A  9     8     |
| 3     14    6     | 15    8     9     | 7     2     145   |
| 57    8     9     | 1257  27    1457  | 6     3     145   |
|-------------------+-------------------+-------------------|
| 1     39    8     | 56    4     2     | 35    7     569   |
| 49    5     7     | 1368  36    18    | 2     146   1469  |
| 2     6     34    | 157   9     157   | 8     145   1345  |
*-----------------------------------------------------------*
A. r1c1 = 4 ==> r2c1 = 6 ==> r2c8 = 4 ==> r4c7 = 4 ==> r1c7 = 1
B. r1c1 = 4 ==> r1c3 = 1

So clearly r1c1 <> 4, and we must have r1c1 = 6.

Since I'm trying to gain a better understanding of the "ALS XZ rule", I went back and tried to cast this exclusion in those terms. This is what I found.

Code: Select all

*-----------------------------------------------------------*
| 46*   7     14+   | 9     25    3     | 145+  8     2456  |
| 469-  39-   5     | 278   1     78    | 34    46    27    |
| 8     2     13+   | 4     57    6     | 9     15    1357  |
|-------------------+-------------------+-------------------|
| 57    14    2     | 36    36    1457  | 145   9     8     |
| 3     14    6     | 15    8     9     | 7     2     145   |
| 57    8     9     | 1257  27    1457  | 6     3     145   |
|-------------------+-------------------+-------------------|
| 1     39    8     | 56    4     2     | 35    7     569   |
| 49-   5     7     | 1368  36    18    | 2     146   1469  |
| 2     6     34    | 157   9     157   | 8     145   1345  |
*-----------------------------------------------------------*
So the only candidates for {r2c1, r2c2, r8c1} are v=3469, the only candidates for {r1c3, r1c7, r3c3} are v=1345, and "3" is restricted common. Since r1c1 can "see" all the instances of "4" in both sets, the rule says I can exclude "4" from r1c1. Everything seems dandy.

But then I got to thinking about the proof that lies behind the "ALS XZ rule". In this instance it doesn't really apply, because I can complete the two ALS's as follows.

If r1c1 = 4 I can set r8c1 = 9, r2c1 = 6, r2c2 = 9
Also, r1c3 = 1, r3c3 = 3, r1c7 = 5

and no contradiction is yet apparent, because the "locked common" value, "3", can "escape" into r7c2. I have to extend the argument farther, along the lines identified as a "DIC" above, before the contradiction is forced.

I'm thinking the reason the proof fails here is that the definition
An almost locked set consists of N cells with exactly N+1 candidates
is not quite restrictive enough. I'm not quite sure how it should be phrased, but it seems that cases like this -- where the two instances of candidate "9" in the set marked with a minus sign can't "see" each other -- need to be excluded somehow. dcb
Ruud
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Post by Ruud »

You are correct, David.

For cells to form an ALS, they need to see each other, just like a naked or hidden subset. For this reason, r2c1, r2c2, r8c1 cannot be an ALS.

Having said that, there are some developments towards extended sets which partially overlap. Myth might be able to tell you more about them, or you could have a look at the Pappocom forum. Search for "COALS".

Ruud
“If the human brain were so simple that we could understand it, we would be so simple that we couldn't.” - Emerson M Pugh
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