The New Sudoku Players' Forum

[SainsburyR]

I have got to the following point and cannot find any logic to go further. Is there some, or have I got to make a guess?

2 8 * 5 * * 6 1 9
9 4 * * * * 8 5 7
* 5 * 9 * 8 3 2 4

1 * 9 * * * 7 8 *
8 * * * 9 * * * *
* 3 4 * * * 1 9 2

3 9 * 6 * 2 * 7 1
4 1 * * * 9 2 3 *
* 7 2 * * 3 9 * 8

[simes]

No need to guess.

Assuming you've done all the easier candidate elimination, you could look at forcing chains. Take a look at either candidate for r3c1, and the effect in r5c8 amongst others.

[tso]

The given position of the board:

Code: Select all

 2 8 . | 5 . . | 6 1 9
 9 4 . | . . . | 8 5 7
 . 5 . | 9 . 8 | 3 2 4
-------+-------+------
 1 . 9 | . . . | 7 8 .
 8 . . | . 9 . | . . .
 . 3 4 | . . . | 1 9 2
-------+-------+------
 3 9 . | 6 . 2 | . 7 1
 4 1 . | . . 9 | 2 3 .
 . 7 2 | . . 3 | 9 . 8



The canditates in each cell can be reduced to the following:

Code: Select all

 (2)    (8)    (37)   (5)    (347)  (47)   (6)    (1)    (9)   
 (9)    (4)    (136)  (123)  (1236) (16)   (8)    (5)    (7)   
*(67)*  (5)    (167)  (9)    (167)  (8)    (3)    (2)    (4)   
 (1)    (26)   (9)    (234)  (234)  (45)   (7)    (8)    (356) 
 (8)    (26)  *(57)*  (123)  (9)    (157) *(45)* *(46)*  (356) 
*(57)*  (3)    (4)    (78)   (678)  (567)  (1)    (9)    (2)   
 (3)    (9)    (58)   (6)    (458)  (2)    (45)   (7)    (1)   
 (4)    (1)    (568)  (78)   (578)  (9)    (2)    (3)    (56)   
*(56)*  (7)    (2)    (14)   (145)  (3)    (9)   *(46)*  (8)   


Notice the cells with *asterisks*.

Code: Select all

r3c1=7 => r6c1=5 => r5c3=7
r3c1=6 => r9c1<>6 => r9c8=6 => r5c8=4 => r5c7=5 => r5c3=7
Therefore, r5c3=7


The rest should be easy.

[SainsburyR]

Thanks tso.

I thought I was using trial & error, when all the time I was using logic.