The New Sudoku Players' Forum

[BadCujo]

I am stumped on this one. Got this far and am drawing blanks.

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

x 7 x 4 x 9 1 8 x
9 x 8 x 2 1 7 x 4
x 1 x 8 7 3 x 5 x

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

[Animator]

Ok, first fix the number 1 in box1.

Then look at row 7, you should be able to see a pair (remove the numbers as candidate from the other cells).
Then look at column 2, again, you should be able to see a pair

Take yet another look at column 2, and you should be able to see a group of 3 numbers... (If you have trouble seeing this one, then imagine r7c3 is empty)

Now after removing them as candidates from the other cells, then you should be able to fill in the value 5 in row 5.

(Also, if you post a grid, can you please include the category?)

[BadCujo]

Thanks Animator. I did not see the pair in the 346 356 at first. Grid went quick after that. Thanks again

[Guest]

I've just discovered the x-wing manoever

You can find a description of it here http://www.simes.clara.co.uk/programs/sudokutechnique6.htm


For your puzzle:

Look at rows 4 and 6, columns 1 and 9.

The x-wing manoever says that because 2 is a posibility in all of them, it can't be in the final solution.

Does that make sense?

[Animator]

That is not an X-wing...

An X-wing requires that there are only two possible cells in the row (or column). (which is the case in Simes's explenation)

The number 2 has these number of possibilities:(without any other elimenation):
* row 4: 3 possible cells
* row 6: 4 possible cells
* column 1: 5 possible cells
* column 9: 4 possible cells

I suggest you look carefully at the explaining grid, and write down where the number 9 can go. You will see that it has only 2 possible cells on row 1 and on row 9.

[BadCujo]

I thought an X-wing had to be only two potential spots in matching row column. I showed a 2 also for r4c3, r6c3, and r6c7. I guess I don't understand X-wing correctly.

[BadCujo]

Maybe I do understand X-wing then.
Was the 5 in r2c3, r4c3, r2c5, r4c5, an X-wing?
I used that to eliminate the 5 in r2c2

[Guest]

ok.

Then what did I do? It worked, or was that a fluke?

[Animator]

At what time? In the starting grid there was not an X-wing in the number 5 either...

In the proces of solving it, there might have been one, but it was not required to solve it... (as in I didn't use an X-wing when solving)


The next notes are about the starting-grid you posted (after fixing the number 1 in box 1):

Column 3 looks good, but column 5 has three candidate cells (r9c5 aswell), and that's why it's not an X-wing... Or that's atleast the situation from the starting grid... (If you managed to remove 5 as candidate from r9c5 then you have an X-wing)

[BadCujo]

I had removed that 5 (r9c5) due to a matched pair 56 in column 5 rows 2 & 4.

[Animator]

Then what did I do? It worked, or was that a fluke?

I have no idea what you did... Also, when I look back at your original reply it says: "... can't be in the final solution". What exactly did you mean with that?

As in, two of the four cells involved in the X-wing has to hold the number. Also note that due to the X-wing in Simes explenation, you can remove 9 as candidate from these (and only these) cells (the cells in column 2 and column 9): r2c2, r2c9, r7c2, r7c9, r8c2, r8c9.