The New Sudoku Players' Forum

[Neunmalneun]

I am not quite sure, but I think the following theory is valid:

If a cell has the candidates "abc.." and "ab" leads to "b" , "a" can be eliminated from this cell.

Maybe this observation is not a big revelation but it can be useful (and is easy to spot) especially in AUR patterns.

Example (top 1465, Number 608)

Code: Select all

 *-----------*
 |...|..3|.1.|
 |...|8.7|..2|
 |8..|6..|...|
 |---+---+---|
 |...|...|...|
 |.69|...|7..|
 |.2.|..8|.6.|
 |---+---+---|
 |..4|3.2|9..|
 |..1|7..|..4|
 |3.8|9..|...|
 *-----------*

 *--------------------------------------------------------------------*
 | 79     4      267    | 5      29     3      | 68     1      678    |
 | 59     13     56     | 8      149    7      | 3456   3459   2      |
 | 8      13     257    | 6      1249   14     | 345    34579  357    |
 |----------------------+----------------------+----------------------|
 | 1457   8      357    | 124    6      9      | 12345  2345   135    |
 | 14     6      9      | 124    3      5      | 7      248    18     |
 | 145    2      35     | 14     7      8      | 1345   6      9      |
 |----------------------+----------------------+----------------------|
 | 6      57     4      | 3      158    2      | 9      578    1578   |
 | 2      9      1      | 7      58     6      | 358    358    4      |
 | 3      57     8      | 9      145    14     | 1256   257    1567   |
 *--------------------------------------------------------------------*


We know (from Uniqueness rule): If R9C8=57 => R7C8=8. That would lead to R9C8<>5 (due to the naked pair 35 in R8C78). So - no matter if our first assumption was correct - R9C8<>5.

Next step would be possible in the AUR R3C56/R9C56 (14). If R3C5=14 => R9C5=5 (leading to R9C6=4 =>R3C6=1 => R3C5<>1). So R3C5<>1

It does not look logical on first sight, but I think it is. If you have the candidates "abcd" for one cell, and "ab" leads to "b", then "a" cannot be right at all (if "c" or "d" are correct, "a" isn't anyway).

[aeb]

I am not quite sure, but I think the following theory is valid:

If a cell has the candidates "abc.." and "ab" leads to "b" , "a" can be eliminated from this cell.

Yes, if the assumption that a cell is either a or b implies that the cell is b then it cannot be a. (For if it is a, then certainly it is one of a and b, and then it is b, contradiction.)

[ronk]

Next step would be possible in the AUR R3C56/R9C56 (14). If R3C5=14 => R9C5=5 (leading to R9C6=4 =>R3C6=1 => R3C5<>1). So R3C5<>1

It does not look logical on first sight, but I think it is. If you have the candidates "abcd" for one cell, and "ab" leads to "b", then "a" cannot be right at all (if "c" or "d" are correct, "a" isn't anyway).

Great deduction. However, I find your exclusion easier to understand from the BUG-Lite POV:

Code: Select all

 79    4     267   | 5     29    3     | 68    1     678
 59    13    56    | 8     149   7     | 3456  3459  2
 8     13    257   | 6     14+29 14    | 345   34579 357
-------------------+-------------------+-----------------
 1457  8     357   | 124   6     9     | 12345 2345  135
 14    6     9     | 124   3     5     | 7     248   18
 145   2     35    | 14    7     8     | 1345  6     9
-------------------+-------------------+-----------------
 6     57    4     | 3     158   2     | 9     578   1578
 2     9     1     | 7     58    6     | 358   358   4
 3     57    8     | 9     14+5  14    | 1256  257   1567

To avoid a deadly pattern of non-uniqueness, at least one of r3c5=2, r3c5=9, and r9c5=5 must be true. These are shown as "extras" in "BUG-Lite+<extras>" notation above.

Noting that ...
r3c5=2 -> r3c5<>1,
r3c5=9 -> r3c5<>1, and
r9c5=5 -> r7c5=1 -> r3c5<>1 (by using the ALS r78c5=158)
In all cases r3c5<>1, validating the exclusion

For those who prefer a nice loop expression:
(AUR:r3c5=2,9|5=r9c5)-5-(ALS:r8c5=5|1=r7c5)-1-r3c5 implying r3c5<>1

Ron