The New Sudoku Players' Forum

[bennys]

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Puzzle: Menneske.no Very Hard #2256842
+-------+-------+-------+
| . 4 . | . 2 3 | . 6 5 |
| . . 5 | 4 . . | 2 3 . |
| . 2 . | . 1 5 | 8 4 . |
+-------+-------+-------+
| . . 1 | 3 7 . | 4 2 . |
| 4 . . | 2 5 1 | . 7 . |
| 7 . 2 | 9 . 4 | . 5 . |
+-------+-------+-------+
| 2 . . | 5 4 . | 6 . 3 |
| . . . | . . 2 | 7 . 4 |
| 6 . 4 | 1 . 7 | 5 8 2 |
+-------+-------+-------+

+-------------------+-------------------+-------------------+
| 189   4     789   | 78    2     3     | 19    6     5     |
|*18    1678  5     | 4    *689  *689   | 2     3    *17    |
| 39    2     3679  | 67    1     5     | 8     4     79    |
+-------------------+-------------------+-------------------+
| 589   5689  1     | 3     7     68    | 4     2     689   |
| 4     368   3689  | 2     5     1     | 39    7     689   |
| 7     368   2     | 9     68    4     | 13    5     168   |
+-------------------+-------------------+-------------------+
| 2     1789  789   | 5     4     89    | 6     19    3     |
| 13589 13589 389   | 68    3689  2     | 7     19    4     |
| 6     39    4     | 1     39    7     | 5     8     2     |
+-------------------+-------------------+-------------------+

A ={R2C1,R2C5,R2C6,R2C9}
Notice that the only 7 in A is box 3 and the only 6's are in box 2.
so we cant have R3C9=7 and R3C4=6 but R3C9=7 => R3C4=6 so we know that R3C9<>7.

[Carcul]

bennys,

That's a very elegant deduction, wich allows the remaining puzzle to be easily solved. I have noted that your conclusion could also be made through the following (not so elegant) nice loop:

Chain 1: [r2c9]=7=[r2c2]=6=[r3c3]-6-[r3c4]-7-[r3c9]=7=[r2c9] => r2c9 = 7.

Regards, Carcul

[rubylips]

Alternatively:

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Consider the chain r1c3-7-r1c4+<8|6>+r3c4=6=r2c2.
When the cell r2c2 contains the value 7, some other value must occupy the cell r1c3, which means that the value 6 must occupy the cell r2c2 - a contradiction.
Therefore, the cell r2c2 cannot contain the value 7.
- The move r2c2:=7 has been eliminated.
The cell r2c9 is the only candidate for the value 7 in Row 2.


[Bob Hanson]

Here's the simple weak link analysis:

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                    1
                    |
         7--6*..6---A---8,9
         |          |
  (r3c9) 7*.........7


Where there is a short 7-7-6 strong chain.

You can't have two weak links to single almost-locked set,
because only one of A6 or A7 can be FALSE. So BOTH r3c9#7 and
r3c4#6 can be eliminated.

Alternatively, just consider the 67 in r3c4 to be a second
almost-locked set. Then we have the standard

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                    1
                    |
        7-B-6...6---A---8,9
        :           |
 (r3c9) 7*..........7


And we can't have r3c9#7 because the two almost-locked sets
are already linked by 6.

Or, using X-type almost-locked sets, we would say:

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                    1
                    |
        7...6*..6---A---8,9
       B|           |
 (r3c9) 7...........7


The 7-7 in row 3 constitutes an almost-locked row set (two column
"candidates" for 7 in the single row 3, like two candidates for a single cell),
thus linking A and B. The 6 at r3c4 would act as a second link to
both and thus can be eliminated.

Basically, this is just plugging almost-locked sets into standard
chain analysis, which is based on the simplest almost-locked sets,
anyway. So ANY chain analysis can be conceptualized in terms of
almost-locked sets. There is an infinite variety of such combinations.

Very interesting.

[bennys]

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I think I wrote before that almost locked sets behave like "directional cells"
what I mean by that is if you have a cell that have two candidates (12)
then this cell is "saying" to every cell that "see" it if any one of you will be 1 then I will eliminate 2 from all of you.
A ALS say less but potentially for more candidates.
It say if I will see 1 in a specific direction then I will eliminate 2 in another specific direction 3 in another and so on.

      +-------------------+     2  3
      | .     1234  234   |----------
      | .     14    .     |
      | .     .     .     |
      +-------------------+
              |
              |1
              |

[ronk]

Code: Select all

Puzzle: Menneske.no Very Hard #2256842
+-------------------+-------------------+-------------------+
| 189   4     789   | 78    2     3     | 19    6     5     |
|*18    1678  5     | 4    *689  *689   | 2     3    *17    |
| 39    2     3679  | 67    1     5     | 8     4     79    |
+-------------------+-------------------+-------------------+
| 589   5689  1     | 3     7     68    | 4     2     689   |
| 4     368   3689  | 2     5     1     | 39    7     689   |
| 7     368   2     | 9     68    4     | 13    5     168   |
+-------------------+-------------------+-------------------+
| 2     1789  789   | 5     4     89    | 6     19    3     |
| 13589 13589 389   | 68    3689  2     | 7     19    4     |
| 6     39    4     | 1     39    7     | 5     8     2     |
+-------------------+-------------------+-------------------+

A ={R2C1,R2C5,R2C6,R2C9}
Notice that the only 7 in A is box 3 and the only 6's are in box 2.
so we cant have R3C9=7 and R3C4=6 but R3C9=7 => R3C4=6 so we know that R3C9<>7.

You obviously posted this puzzle before you refined your Almost Locked Set xz rule. If you did so today, it might look like ...

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+-------------------+-------------------+-------------------+
| 189   4     789   | 78    2     3     | 19    6     5     |
|*18    1678  5     | 4    *689  *689   | 2     3    *17    |
| 39    2     3679  |^67    1     5     | 8     4     79    |
+-------------------+-------------------+-------------------+
| 589   5689  1     | 3     7     68    | 4     2     689   |
| 4     368   3689  | 2     5     1     | 39    7     689   |
| 7     368   2     | 9     68    4     | 13    5     168   |
+-------------------+-------------------+-------------------+
| 2     1789  789   | 5     4     89    | 6     19    3     |
| 13589 13589 389   | 68    3689  2     | 7     19    4     |
| 6     39    4     | 1     39    7     | 5     8     2     |
+-------------------+-------------------+-------------------+

A = {R2C1,R2C5,R2C6,R2C9}
B = {R3C4}
x = 6
z = 7

... permitting safe elimination of z in the intersection of row 3 and box 3, and the intersection of row 2 and box 2 (excluding A and B, of course).

Therefore, I think that Bob Hanson's "formal identification" of R3C4 as a set was proper.