The New Sudoku Players' Forum

[ArkieTech]

Thanks SteveC -- most helpful

does this say it?

[(78=15)r39c19]-1r25c1; ste

I didn't think about a chain in this particular CoALS elimination.
However, something should include the contradiction cell (78)r3c5. I'll work on it!

SteveC

[(78=15)r39c19,r3c5]-1r25c1; ste

[Sudtyro2]

Code: Select all

 *--------------------------------------------------*
 | 3    4    7    | 1    2    6    | 9    5    8    |
 | 68-1 9    16   | 5    3    78   | 17   2    4    |
 | 158a 2    15   | 4    78   9    | 17   3    6    |
 |----------------+----------------+----------------|
 | 2    5    9    | 8    1    3    | 6    4    7    |
 | 6-1  16   8    | 7    4    2    | 3    9    5    |
 | 7    3    4    | 6    9    5    | 8    1    2    |
 |----------------+----------------+----------------|
 | 9    7    3    | 2    6    4    | 5    8    1    |
 | 4    16   156  | 9    58   18   | 2    7    3    |
 | 15ab 8    2    | 3    57b  17   | 4    6    9    |
 *--------------------------------------------------*

Dan, maybe this short stream shows how the contradiction cell, r3c5, can be explicitly included. I've added in logical operators for clarity...

Code: Select all

(7|8)r3c5 – (7&8=5&1)r39c1,r9c5 – (1)r25c1

In words, it reads something like:
Digit 7 or 8 must be true in r3c5, so digits 7 and 8 cannot both be true in the combined ALS. Thus digits 1 and 5 must be true in the combined ALS, where digits (1)r39c1 are both seen by external digits (1)r25c1. Hence, r25c1<>1.

[Edit to fix typo r9c5]
SteveC

[ArkieTech]

Code: Select all

 *--------------------------------------------------*
 | 3    4    7    | 1    2    6    | 9    5    8    |
 | 68-1 9    16   | 5    3    78   | 17   2    4    |
 | 158a 2    15   | 4    78   9    | 17   3    6    |
 |----------------+----------------+----------------|
 | 2    5    9    | 8    1    3    | 6    4    7    |
 | 6-1  16   8    | 7    4    2    | 3    9    5    |
 | 7    3    4    | 6    9    5    | 8    1    2    |
 |----------------+----------------+----------------|
 | 9    7    3    | 2    6    4    | 5    8    1    |
 | 4    16   156  | 9    58   18   | 2    7    3    |
 | 15ab 8    2    | 3    57b  17   | 4    6    9    |
 *--------------------------------------------------*

Dan, maybe this short stream shows how the contradiction cell, r3c5, can be explicitly included. I've added in logical operators for clarity...

Code: Select all

(7|8)r3c5 – (7&8=5&1)r39c1,r9c3 – (1)r25c1

In words, it reads something like:
Digit 7 or 8 must be true in r3c5, so digits 7 and 8 cannot both be true in the combined ALS. Thus digits 1 and 5 must be true in the combined ALS, where digits (1)r39c1 are both seen by external digits (1)r25c1. Hence, r25c1<>1.

SteveC

Thanks for putting up with a slow learner.