The New Sudoku Players' Forum

[speter]

G'Day Folks,

I was solving the puzzle below and noticed the pattern with the 4's in rows 2 & 3.
Since either r2c3 or r3c3 must be 4; I can discard any 4's that can see them.



My question is: is this considered a swordfish (and if not, what is it)?

cheers
S.

[Leren]

Can't see an obvious fish on 4 there. The puzzle solves easily with a lot of intersections, that I've marked up on this PM. The puzzle solves with singles from there. Leren

Code: Select all

*-----------------------------------------------------*
| 456   156     7     | 9  3   48  |  2     15   1468 |
| 2     35      3458  | 1  6   7   |  89-5  59   489  |
| 9     16      1468  | 48 5   2   |  1678  17   3    |
|---------------------+------------+------------------|
|*357   8      *359   | 6  79  1   |#*359   4    2    |
| 1     2      *59-3  |*38 4  *38  | #59-3  6    7    |
|*367   4      *369   | 2  79  5   | *139   8    19   |
|---------------------+------------+------------------|
| 8     1679-3 16-349 |*34 2  *346 |  13679 1379 5    |
|*3456 *356    *3456  | 7  1   9   |  68-3  2    68   |
| 6-37  1679-3  2     | 5  8   36  |  4     1379 169  |
*-----------------------------------------------------*

[yzfwsf]

Sashimi Swordfish:4r238\c349 fr8c1 => r7c3<>4

[Mauriès Robert]

Hi speter,

My question is: is this considered a swordfish (and if not, what is it)?

This pattern yellow is a Skyscraper, but the eliminations you make in red are not due to this pattern.
Note that 4r7c3 is eliminated due to the 4r8B7 or 4r7B8 alignment.

Since either r2c3 or r3c3 must be 4; I can discard any 4's that can see them.

How do you explain that r2c3 or r3c3 is necessarily 4?
Cordialy
Robert

[speter]

How do you explain that r2c3 or r3c3 is necessarily 4?

Yes, you are correct.

I was incorrectly thinking that since r2c3 or r2c9 must be 4; and r3c3 or r3c4 must be 4; then r2c3 or r3c3 must be 4.

Thanks for pointing out my error.

cheers
S.

[Mauriès Robert]

Hi speter,

How do you explain that r2c3 or r3c3 is necessarily 4?

Yes, you are correct.
I was incorrectly thinking that since r2c3 or r2c9 must be 4; and r3c3 or r3c4 must be 4; then r2c3 or r3c3 must be 4.
Thanks for pointing out my error.

If your reasoning is wrong, the result is right. One can easily show, with a chain, that r2c3 or r3c3 is 4, like this :
(-4r2c23) => 4r1c1->8r1c6->3r5c6->3r7c4->6r9c6->3r9c1 => -3B9 which is impossible, so 4 is r2c2 or r2c3.
From then on we can make the eliminations you indicate.
That said, the basic techniques are sufficient to solve this puzzle without these eliminations, simply by noticing the alignments, especially the 3s and 7s and the 36r9c16 doublet that eliminates the 3r9c8. The puzzle then ends with singles.
Cordialy
Robert