The New Sudoku Players' Forum

[bennys]

Code: Select all

Puzzle: Menneske.no Very Hard #1915887
+-------+-------+-------+
| 5 . . | 8 7 2 | . 1 . |
| . . . | 6 1 5 | . . 7 |
| . 1 . | 3 9 4 | . . . |
+-------+-------+-------+
| 9 7 6 | 1 4 3 | 8 2 5 |
| 4 5 3 | 2 8 6 | 9 7 1 |
| 2 8 1 | 7 5 9 | 6 4 3 |
+-------+-------+-------+
| . . . | 5 6 1 | . . . |
| 1 . . | 4 2 8 | . . . |
| . 2 . | 9 3 7 | 1 . 4 |
+-------+-------+-------+




+----------------+----------------+----------------+
| 5    3469 49   | 8    7    2    | 34   1    69   |
| 38   349  2489 | 6    1    5    | 234  389  7    |
| 678  1    278  | 3    9    4    | 25   568  28   |
+----------------+----------------+----------------+
| 9    7    6    | 1    4    3    | 8    2    5    |
| 4    5    3    | 2    8    6    | 9    7    1    |
| 2    8    1    | 7    5    9    | 6    4    3    |
+----------------+----------------+----------------+
| 378  349  4789 | 5    6    1    | 27   389  28   |
| 1    369  579  | 4    2    8    | 357  359  69   |
| 68   2    58   | 9    3    7    | 1    568  4    |
+----------------+----------------+----------------+


It can't be that R1C2,R1C3 and R2C2 are all 4,9
So we have in that set 3 or 6
That mean that in R2C1,R3C1 we cant have both 6 and 3
Which mean that or we don't have 6 there
and then R9C1 = 6 or we don't have 3 (R2C1=8) and we get R9C1 = 6 again.

[Jeff]

Nice deduction. This is no ordinary deduction because it involves a forcing net with 4 implications.

r1c2=6 => r3c1<>6 => r9c1=6
r1c2=3 => r2c1=8 => r9c1=6
{r1c2=4 => r1c3=9} => r2c2=3 => r2c1=8 => r9c1=6
{r1c2=9 => r1c3=4} => r2c2=3 => r2c1=8 => r9c1=6
Therefore r9c1=6

Another deduction in this grid can be obtained via the following double implication chain.

[r2c3]=2=[r3c3]=7=[r3c1]=6=[r1c2]=3=[r1c7]=4=[r2c7]=2=[r2c3] => r2c3=2

[rubylips]

The principle here is the variation on Two Sector Disjoint Subsets that cropped up in DanO's 'More Forcing Chains' puzzle. Its most general form is as follows:

Define an Almost Disjoint Subset to be a set of n cells populated by n+1 distinct values.

Suppose we have a Row (or Column) and a Box, such that the Row contains an Almost Disjoint Subset in the cells outside the Box and the Box contains an Almost Disjoint Subset in the cells outside the Row. Furthermore, suppose that within the intersection there is a cell with two candidate values, each of which appears in just one of the Almost Disjoint Subsets. Then any value that is common to the Almost Disjoint Subsets should be eliminated from the intersection.

That's a bit wordy, so here are two illustrative examples:

Code: Select all

+----------------+----------------+----------------+
| 5    3469 49   | 8    7    2    | 34   1    69   |
| 38   349  2489 | 6    1    5    | 234  389  7    |
| 678  1    278  | 3    9    4    | 25   568  28   |
+----------------+----------------+----------------+
| 9    7    6    | 1    4    3    | 8    2    5    |
| 4    5    3    | 2    8    6    | 9    7    1    |
| 2    8    1    | 7    5    9    | 6    4    3    |
+----------------+----------------+----------------+
| 378  349  4789 | 5    6    1    | 27   389  28   |
| 1    369  579  | 4    2    8    | 357  359  69   |
| 68   2    58   | 9    3    7    | 1    568  4    |
+----------------+----------------+----------------+

Box 1 has an ADS in r1c2, r1c3 and r2c2 with values {3,4,6,9}.
Column 1 has an ADS in r9c1 with values{6,8}.
The cell r2c1 has two candidates, one from each ADS.
Therefore, whichever value populates r2c1, one of the two Subsets will be completed and the value 6 will be accounted for. It follows that 6 cannot be a candidate for r3c1, which leaves r9c1 as the sole location for a 6 in Column 1.

Code: Select all

+-------------------+-------------------+-------------------+
| 235   359   2359  | 7     14    59    | 8     14    6     |
| 57    6     8     | 3     14    59    | 149   2     17    |
| 4     79    1     | 8     6     2     | 39    379   5     |
+-------------------+-------------------+-------------------+
| 1235  1359  2359  | 6     7     8     | 124   145   123   |
| 13    4     6     | 5     2     13    | 7     8     9     |
| 8     1357  2357  | 4     9     13    | 6     15    123   |
+-------------------+-------------------+-------------------+
| 6     135   4     | 29    38    7     | 12359 39<   128   |
| 9     2     357   | 1     38    4     |>35    6    >78    |
| 137   8     37    | 29    5     6     | 129  <379>  4     |
+-------------------+-------------------+-------------------+

Row 8: {3,8} in r8c5.
Box 9: {3,7,9} in r7c8 and r9c8.
Link Cell: {7,8} in r8c9.
Therefore, eliminate 3 from r8c7.

BTW, the first puzzle falls to the following chain:

Code: Select all

Consider the chain r1c7-3-r1c2-6-r1c9-9-r2c8.
When the cell r2c8 contains the value 3, some other value must occupy the cell r1c7, which means that the value 9 must occupy the cell r2c8 - a contradiction.
Therefore, the cell r2c8 cannot contain the value 3.
- The move r2c8:=3 has been eliminated.
The value 3 in Box 9 must lie in Column 8.
- The move r8c7:=3 has been eliminated.
The values 3 and 4 occupy the cells r1c7 and r2c7 in some order.
- The move r2c7:=2 has been eliminated.
The cell r2c3 is the only candidate for the value 2 in Row 2.


[bennys]

I used the same concept but gave it different name here http://forum.enjoysudoku.com/viewtopic.php?p=12244&highlight=#12244.
But you are right it looks like I am using it here.
I think this thing is giving us an OR statements and when combining a few of them (i mean OR statements) we get results.
For example something like that.

Code: Select all

+----------------+----------------+----------------+
| 5    3469 49   | 8    7    2    |                |
| 38   349  2489 | 6    1    5    | 78             |
| 678  1    278  | 3    9    4    | 76             |
+----------------+----------------+----------------+


Here again we can eliminate the 6 from R3C1.

[rubylips]

My solver would approach your earlier problem as follows:
It would identify the chain X-4-Z-7-W-5-Y.
It would recognize that X<>4 => Y=5.
It would immediately eliminate 4 from Y - it wouldn't have to consider the values 1, 2 or 3 - since if Y were to contain 4, X would have to contain something else and Y would contain 5 - a contradiction. This type of logic appears in the solver extract above.
Next, it would recognize that Y<>5 => X=4, which means that X & Y must bewteen them contain exactly one of 4 or 5. In turn this means that they must contain exactly one value from 1, 2 and 3 - which completes the Disjoint Subset, and allows any further instance of 1, 2 or 3 to be eliminated from the row.

[Jeff]

Post edited and moved to thread "stuck" originated by MikeF.