The New Sudoku Players' Forum

[marek stefanik]

We can also use a parity argument:

Code: Select all

.------------------.-----------------.---------------------.
|#158    2    3    | 4  B#158   69   | 678  1568    15679  |
| 1458  #158  7    |C#158 1358  69   | 2    134568  134569 |
| 6      9    1458 | 2    7    A1358 | 348  13458   1345   |
:------------------+-----------------+---------------------:
| 2458   358  2458 | 7    9     2358 | 1    346     346    |
| 1279   13   129  | 6    4     123  | 5    37      8      |
| 14578  6    1458 | 158  1358  1358 | 9    347     2      |
:------------------+-----------------+---------------------:
| 3      4   A158  | 9    6   A#158  | 78   2       157    |
|C#158   7    6    |C#158 2     4    | 38   9       135    |
| 29   B#158  29   | 3  B#158   7    | 468  14568   1456   |
'------------------'-----------------'---------------------'

Suppose -3r3c6.
Then the permutations of the ABC-marked cells (in order) have the same parity in each of b278.
(If b28 had different parity, one digit would appear twice in the same column, same with b78, one digit would appear twice in one row. Having the same parity is an equivalence relation.)
Suppose b27 have the same permutation, ie. their respective A, B, and C cells contain the same digits, say abc in this order.
Then b1p15 are seen by both b and c and they must therefore both be a, ie. contra.
Therefore b27 have different permutations of the same parity and their respective A, B, and C cells all contain a pair of different digits.
Together with the trivial weak links, each digit can only appear once in each of ABC, making them remote triples.

Marek

[DEFISE]

Hi Cenoman (and others)

It's Ok, I made my little proof. We just need to show that r3c6 and r7c3 cannot be equal. The depth of proof is 3 but it is still very short.
Very nice trick this RT !