The New Sudoku Players' Forum

[sudokuer]

I AM STUCK ON THIS PUZZLE ... ANY IDEA WHAT THE NEXT STRATEGY IS?

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


_ _ 2 4 7 _ 9 5 _
5 9 _ 2 _ 1 8 7 4
_ 7 4 _ 5 9 2 1 _


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

[tarek]

I can't tell where you've actually reached in your solution without actual pancilmarks....

But assuming you managed some box-line interaction eliminations...

Code: Select all

*-----------------------------------------------*
| 6    3    1   | 7    9    5   | 4    8    2   |
| 9    8    5   | 1    4    2   | 6    3    7   |
| 4    2    7   | 68   36   38  | 5    9    1   |
|---------------+---------------+---------------|
| 138  16   2   | 4    7   *38  | 9    5   *36  |
| 5    9    36  | 2    36   1   | 8    7    4   |
| 38   7    4   |*68   5    9   | 2    1   -36  |
|---------------+---------------+---------------|
| 2    16   68  | 9    18   7   | 3    4    5   |
| 13   4    38  | 5    18   6   | 7    2    9   |
| 7    5    9   | 3    2    4   | 1    6    8   |
*-----------------------------------------------*
Eliminating 6 From r6c9 (3 & 8 in r4c6 form an XY wing with 6 in r4c9 & r6c4)

There is a sticky thread on this forum explaining how to post puzzles & pencilmarks, here is a link to it http://forum.enjoysudoku.com/viewtopic.php?t=2664

Tarek

[QBasicMac]

Got it? If not, then in other words:

If r4c9=3 then f6c9=6. That leave only 8 in r4c6 AND r6c4 = impossible.
That means r4c9 must be 6, which solves the puzzle.

Mac

[tarek]

hi QBasicMac,

I don't think that your method equates with the XY wing mentioned above...

r4c6=3 => r4c9=6 => r6c9<>6=> r6c9=3
r4c6=8 => r6c4=6 => r6c9<>6=> r6c9=3
Therefore Eliminating 6 from r6c9 leaving 3.

That is how it works......

Tarek

[ronk]

I don't think that your method equates with the XY wing mentioned above...

QBasicMac described a "backtest" method for the xy-wing deduction, the very backtest illustrated here. "Backtest" is defined here.

QBasicMac used it as a "proof by contradiction" ... so to speak ... in lieu of a forcing chain.

Ron

[tarek]

QBasicMac used it as a "proof by contradiction" ... so to speak ... in lieu of a forcing chain.

That clears it, thanx....

Tarek

[tso]

The easiest solution from the position Tarek gave is BUG:

Code: Select all

*-----------------------------------------------*
| 6    3    1   | 7    9    5   | 4    8    2   |
| 9    8    5   | 1    4    2   | 6    3    7   |
| 4    2    7   | 68   36   38  | 5    9    1   |
|---------------+---------------+---------------|
| 138  16   2   | 4    7    38  | 9    5    36  |
| 5    9    36  | 2    36   1   | 8    7    4   |
| 38   7    4   | 68   5    9   | 2    1    36  |
|---------------+---------------+---------------|
| 2    16   68  | 9    18   7   | 3    4    5   |
| 13   4    38  | 5    18   6   | 7    2    9   |
| 7    5    9   | 3    2    4   | 1    6    8   |
*-----------------------------------------------*


r4c1 is the only polyvalue cell.

The candidate 3 is the only one that appears three times in the same row, column and box as r4c1.

r4c1 must be 3 to avoid BUG.