The New Sudoku Players' Forum

[Fortesque92]

https://postimg.cc/wRmvkNb8

Found an empty rectangle in box 9 but then got stuck.

[jco]

Givens:

Code: Select all

*-----------*
 |.6.|9..|..8|
 |.1.|.2.|...|
 |.9.|.38|.2.|
 |---+---+---|
 |4..|...|...|
 |6..|.82|.94|
 |2..|...|18.|
 |---+---+---|
 |...|.4.|.3.|
 |..8|..9|.6.|
 |..5|...|.42|
 *-----------*

After basics: look at the pattern of 7 (the mentioned ER is not needed, but it does no harm).

[Fortesque92]

Sorry I didn't quite understand that, could you elaborate?

[jco]

Sorry I didn't quite understand that, could you elaborate?

In your current configuration either 7 r1c5 or 7 r5c4 is true. If you prove this, then you can
remove 7 from r46c5.
Another way to solve is to look for a unique rectangle at blocks 4 and 5.

[Leren]

There is a relatively painless way to solve this puzzle, that Hodoku chose. It's called Remote Pairs.

Code: Select all

*---------------------------------------------*
|A57 6  2  | 9    B57     4     | 3   1   8   |
| 8  1  3  | 567   2      567   | 4  B57  9   |
|B57 9  4  | 1     3      8     | 6   2  A57  |
|----------+--------------------+-------------|
| 4  8  79 | 3567  169-57 13567 | 2  A57  36  |
| 6  35 1  | 357   8      2     |B57  9   4   |
| 2  35 79 | 4     5679   3567  | 1   8   36  |
|----------+--------------------+-------------|
| 19 2  6  | 578   4      157   | 89  3   157 |
| 3  4  8  | 2     1-57   9     |A57  6   157 |
| 19 7  5  | 368   16     136   | 89  4   2   |
*---------------------------------------------*

After basics, there a lot of cells with 57 in them. I've labelled them A & B. What this indicates is the parity of the 57 cells.

If one 57 cell labelled A is 5, then all the 57 A cells are 5 and the 57 B cells are 7. Similarly if If one 57 cell labelled A is 7, then all the 57 A cells are 7 and the 57 B cells are 5.

The clever thing about this arrangement is that any cell that can see an A and a B 57 cell can't be 57, of which I've indicated two in the PM.

A more complex move than I would normally recommend but it does solve the puzzle in one non-basic move.

You can read up on Remote Pairs on Hodoku here.

Leren

[Fortesque92]

Mind blowing, cheers!

[jco]

There is a relatively painless way to solve this puzzle, that Hodoku chose.
(...)
Leren

Nice! I did see that remote pair (eliminations) but I did not realize they would be enough to finish the puzzle in one step.
So, I looked for something else. My one-stepper is more complex (some fish?):

Code: Select all

.---------------------------------------------------------------------.
| 57     6      2      | 9     a5(7)   4      |  3      1      8      |
| 8      1      3      |b56(7)  2     b56(7)  |  4     c5(7)   9      |
| 57     9      4      | 1      3      8      |  6      2      57     |
|----------------------+----------------------+-----------------------|
| 4      8      79     | 3567   1569-7 13567  |  2     d5(7)   36     |
| 6      35     1      |f35(7)  8      2      | e5(7)   9      4      |
| 2      35     79     | 4      569-7  3567   |  1      8      36     |
|----------------------+----------------------+-----------------------|
| 19     2      6      | 578    4      157    |  89     3      157    |
| 3      4      8      | 2      15-7   9      |f'5(7)   6      157    |
| 19     7      5      | 368    16     136    |  89     4      2      |
'---------------------------------------------------------------------'

(7): r1c5 = r2c46 - r2c8 = r4c8 - r5c7 = r5c4 & r8c7 => -7 r468c5; ste

A simple way to describe: if we don't have (7)r1c5, the we must have (7)r2c46, so no (7)r2c8,
which in turn implies that we must have (7)r4c8, and so we can't have (7)r5c4, and this
finally implies that we must have (7)r5c4 and (7)r8c7. Thus, either (7)r1c5 or [7)r5c4 and (7)r8c7]

Btw, I did not see ER at block 9 [perhaps another move?], but the eliminations were correct
[I mean this regarding the configuration in the link]