The New Sudoku Players' Forum

[TheDonald]

Help! What is my Next Move???
Given Grid:
x x 2 | 5 x x | 6 x x
1 x x | x x x | x x x
x x x | 6 x 8 | x 1 x

x 2 x | 1 x x | x 9 x
x x 9 | x x x | x x x
x 3 1 | x x x | x x 7

6 x 7 | x x 2 | 4 3 1
x x x | x x x | 9 x x
x 4 x | 3 7 x | x x x

What I have So Far Is This:
x x 2 | 5 1 x | 6 x x
1 x x | x x x | x x x
x x x | 6 x 8 | x 1 x

x 2 x | 1 x x | x 9 x
x x 9 | x x x | 1 x x
x 3 1 | x x x | x x 7

6 x 7 | x x 2 | 4 3 1
2 1 3 | x x x | 9 7 x
9 4 x | 3 7 1 | x x x

[Shazbot]

My next move would be to give up - this puzzle has 710 solutions! Sorry

[Animator]

My next move would be to give up - this puzzle has 710 solutions! Sorry :(

This is not correct.

As the title suggest: this is a Sudoku X puzzle. (The diagonals need to have the numbers 1-9 aswell).

There is a unique solution.

TheDonald: Did you read New topics - How to ask for help? I'm guessing that it would need to be in the different forum (unless you got a program that generated it) and the layout of the grid could be made even more readable...

[Animator]

The next move:

Take a really good look at the number 6.


Where can it go in the second diagonal (= upper right to bottom left)? Only in box 5.
This means: remove 6 as a candidate from: r4c5, r5c6, r6c5, r6c6.

This leaves the candidate 6 in: r4c6 and r5c5.
Now take a look at box 3. 6 is possible in: r4c3 and r5c2.

This means that the number 6 on row 6 has to be in box 6.

Fill it in and you have your next number.


Another option: take a good look at the number 7.

The second diagonal already has the number 7. This means you can remove 7 as a candidate from r3c7. This leaves only once cell as a candidate for the number 7 in box 3.

Update: added the second option.

[Shazbot]

Oops - sorry. I've never heard of a Sudoku-X before! Just dubbed into Simple Sudoku to see the grid, and that's what it came up with. I'll watch out for these in future and leave the solving to the Sudoku-X - Xperts!

[QBasicMac]

Sudoku-X - Xperts!

Evidently not me. First time, too.

Well, I solved as far as I could and then noted some placements on the diagonals, so I eliminated pencilmarks there and was able to get to here:

Code: Select all

+-------------+------------------+------------------+
| 38   9   2  | 5   1       7    | 6    48    34    |
| 1    68  46 | 29  2349    349  | 7    2458  23459 |
| 34   7   5  | 6   2349    8    | 23   1     2349  |
+-------------+------------------+------------------+
| 7    2   46 | 1   34568   3456 | 358  9     3456  |
| 458  68  9  | 7   2368    3456 | 1    2456  23456 |
| 458  3   1  | 28  245689  69   | 258  2456  7     |
+-------------+------------------+------------------+
| 6    5   7  | 89  89      2    | 4    3     1     |
| 2    1   3  | 4   56      56   | 9    7     8     |
| 9    4   8  | 3   7       1    | 25   256   26    |
+-------------+------------------+------------------+


Now I see that r6c6 has the only candidate 9 on diag r1c1 so I made that placement. Then I saw a locked candidate 8 in box 1 which eliminates one 8 in r5c5

Then I gave up. But that is the basic path, I think.

Mac

[QBasicMac]

Well, more progress my way:

By T&E, it r9c9 cannot be 2. It leads to 2 in r6c4 and r3c7 (diagonal)
Therefore r9c9=6

Eliminating 6's on the diagonal give r2c2=8

Solving some singles leads to

Code: Select all

+---------+------------+----------+
| 3  9  2 | 5  1   7   | 6  8  4  |
| 1  8  6 | 2  3   4   | 7  5  9  |
| 4  7  5 | 6  9   8   | 3  1  2  |
+---------+------------+----------+
| 7  2  4 | 1  56  356 | 8  9  35 |
| 8  6  9 | 7  2   35  | 1  4  35 |
| 5  3  1 | 8  4   9   | 2  6  7  |
+---------+------------+----------+
| 6  5  7 | 9  8   2   | 4  3  1  |
| 2  1  3 | 4  56  56  | 9  7  8  |
| 9  4  8 | 3  7   1   | 5  2  6  |
+---------+------------+----------+


where we see that r4c6 is the only 5 on that diagonal, which gives the solution immediately.

Mac

[Animator]

Well, more progress my way:

By T&E, it r9c9 cannot be 2. It leads to 2 in r6c4 and r3c7 (diagonal)
Therefore r9c9=6

There really is no need for trial and error.

From your first set of pencilmark: Look at the second diagonal. Does it has the number 6? No.

So either: r1c9, r2c8, r3c7, r4c6, r5c5, r6c4, r7c3, r8c2, r9c1 has to be the number 6.

r1c9, r2c8, r3c7 : impossible, box 3 already has the number 6
r7c3, r8c2, r9c2: those numbers are already filled in.

This leaves us with r4c6 or r5c5. Both of these cells are in box 5.
This means that you can remove 6 as a candidate from: 4c5, r6c5, r7c5, r7c6. Now look at r7c6. It has only one number...

[QBasicMac]

There really is no need for trial and error.

Heh - as some say, there is never need for T&E. But I took your message to mean "There is no need for Boring Arcane Techniques, so no justification for T&E. Actually, I just got lazy and quit looking and wanted to see if the puzzle indeed had a unique solution.

Thanks for the move!! That encouraged me to try again. Below is a path to solution using your move and similar ones I could then spot.

Mac


Ignoring X and solving singles, etc. gets to here

Code: Select all

+-------------------+---------------------+----------------------+
| 3478  789    2    | 5     1       3479  | 6      48     3489   |
| 1     56789  4568 | 2479  2349    3479  | 23578  2458   234589 |
| 3457  579    45   | 6     2349    8     | 2357   1      23459  |
+-------------------+---------------------+----------------------+
| 4578  2      4568 | 1     34568   34567 | 358    9      34568  |
| 4578  5678   9    | 2478  234568  34567 | 1      24568  234568 |
| 458   3      1    | 2489  245689  4569  | 258    24568  7      |
+-------------------+---------------------+----------------------+
| 6     58     7    | 89    589     2     | 4      3      1      |
| 2     1      3    | 48    4568    456   | 9      7      58     |
| 9     4      58   | 3     7       1     | 258    2568   2568   |
+-------------------+---------------------+----------------------+


Now eliminating candidates on the X due to 97 and 47 gets to here

Code: Select all

+------------------+---------------------+----------------------+
| 38    789   2    | 5     1       3479  | 6      48     348    |
| 1     5689  4568 | 2479  2349    3479  | 23578  2458   234589 |
| 3457  579   5    | 6     2349    8     | 235    1      23459  |
+------------------+---------------------+----------------------+
| 4578  2     4568 | 1     34568   3456  | 358    9      34568  |
| 4578  5678  9    | 2478  23568   34567 | 1      24568  234568 |
| 458   3     1    | 248   245689  569   | 258    24568  7      |
+------------------+---------------------+----------------------+
| 6     58    7    | 89    589     2     | 4      3      1      |
| 2     1     3    | 48    4568    456   | 9      7      58     |
| 9     4     58   | 3     7       1     | 258    2568   2568   |
+------------------+---------------------+----------------------+


Ignore X again and solve singles, etc.
Run across r1c2=r1c6=(hidden pair)79 in the process
Finally get to here

Code: Select all

+-------------+------------------+------------------+
| 38   9   2  | 5   1       7    | 6    48    34    |
| 1    68  46 | 29  2349    349  | 7    2458  23459 |
| 34   7   5  | 6   2349    8    | 23   1     2349  |
+-------------+------------------+------------------+
| 7    2   46 | 1   34568   3456 | 358  9     3456  |
| 458  68  9  | 7   23568   3456 | 1    2456  23456 |
| 458  3   1  | 28  245689  569  | 258  2456  7     |
+-------------+------------------+------------------+
| 6    5   7  | 89  89      2    | 4    3     1     |
| 2    1   3  | 4   56      56   | 9    7     8     |
| 9    4   8  | 3   7       1    | 25   256   256   |
+-------------+------------------+------------------+


Note that r4c6 and r5c5 have locked candidate 6 on the diagonal. Erase 6 from all other cells in box 5.
Do a few singles and note that r5c5 and r6c6 are locked candidate 5 on the diagonal. Remove 5's from box 5.
Now r5c5=56(hidden pair with r8c5)
Nice!: r1c1=3 (single on the diagonal)
Solve some more singles and find Nice!: r3c7=3 (single on the diagonal).
That does it!