The New Sudoku Players' Forum

[stuartn]

Found it. After the clever removal of the candidate 3 in R9C4 - and doing all the basic housekeeping - we're left with two candidates (69) at R3C5. There are two chains (one obvious) which force this cell to be a 6. Once this is realised the grid falls apart.

1. R4C5= 9 => R3C5 = 6.

2. R4C5=3 => R7C5=59 => R7C4 = 3 =>R7C6=1 =>R3C6=489 =>R1C6= 89. R7C6= 1 =>R7C7=4 => R8C7 = 69 =>R7C8=8 = >R3C8=69 =>R1C8 = 69 => R2C7=1 and R1C7 = 8 => R1C6=9 => R3C5 = 6

Just ask if more info needed.

Stuartn

[Jeff]

stuartn, what happen if r4c5=7 or 9? BTW, what you have listed is only one chain, but with double implication.

[tso]

If you filter on 2's alone, you'll have no trouble placing all nine 2's.

Not without r2c2 having been determined as 7.

Actually, even with r2c2=7, I can still place all nine 2's -- though it will leave the grid unsolvable. Nishio alone doesn't cover this. When applying Nishio, there are two possible results -- either you CANNOT place all nine digits, in which case you elliminate the possibility from the cell you were testing, or you CAN place them all, in which case you cannot make the ellimination. For example:

Code: Select all

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

All nine 2's are comfortably seated -- though there is nowhere for the last 7. Performing Nishio in your head -- you probably wouldn't consider this. In the puzzle at the top of the thread, even if you hadn't entered any candidates, you could easily do that Nishio in your head.


So you could state it with a proof by contradiction:

1) r6c6=2 => r3c6=7 => r2c2=7 => r9c2=2 => r9c3=7
... and ...
2) r6c6=2 => r7c3=2 => r8c8=2 => r3c7=2 => r9c3=2
This is a contradiction, therefore, r6c6<>2

The second line is Nishio-esque.