The New Sudoku Players' Forum

[Kent]

I solved this puzzle using 3 xy wings.I'm just wondering whether there's any alternative method to solve.

Code: Select all

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


 2    8    4    | 9    6    5    | 3    7    1   
 39   35   59   | 4    7    1    | 8    2    6   
 7    6    1    | 8    3    2    | 9    4    5   
----------------+----------------+----------------
 1    9    2    | 7    4    6    | 5    3    8   
 6    4    8    | 35   59   39   | 7    1    2   
 5    7    3    | 1    2    8    | 6    9    4   
----------------+----------------+----------------
 349  2    579  | 35   8    47   | 1    6    39 
 3489 1    579  | 6    59   47   | 2    58   39 
 389  35   6    | 2    1    39   | 4    58   7   


[Carcul]

Yes, there are certainly several alternative ways of solve that puzzle. Here is one of them:

Almost Locked Sets xz-rule:

A=r7c3/r8c3/r9c2; B=r9c6; x=3 and z=9, => r9c1<>9 which solve the puzzle.

Regards, Carcul

[Kent]

What's almost locked xz-rule??This isn't mentioned much on many of the sites.

[ravel]

What's almost locked xz-rule??This isn't mentioned much on many of the sites.

You can find some ALS-links here

To understand this one:
In set A (r7c3/r8c3/r9c2) you have candidates 3579
In set B (r9c6) 39
When you have n cells in a set, there must be n+1 possible candidates, so, if one candidate would not be possible, a single, pair, triple, quad, etc is left (the set is locked).
There must be 2 common candidates, in this case x=3 and z=9.
z is the candidate you can eliminate outside the sets in one or more cells, that "see" all z's in both sets. x is the candidate, that - after the z is occupied outside, has to go to different cells in a common unit, so you would get a contradiction.
So, if r9c1=9, there is a triple 357 in set A (or a pair 57 in r7c3/r8c3), so 3 has to be in r9c2.
On the other hand in set B only 3 is left in r9c6.
This is a contradiction.

For finding ALS's you need a good eye and some practice. For me personally it is easier to see directly, that a 9 in r9c1 forces both the 3 in r9c6 and the pair 57 in r78c3, which leaves a 3 in r9c2.

[Sped]

Code: Select all

 
*-----------------------------------------------------------*
 | 2     8     4     | 9     6     5     | 3     7     1     |
 | 39    35    59    | 4     7     1     | 8     2     6     |
 | 7     6     1     | 8     3     2     | 9     4     5     |
 |-------------------+-------------------+-------------------|
 | 1     9     2     | 7     4     6     | 5     3     8     |
 | 6     4     8     | 35    59    39    | 7     1     2     |
 | 5     7     3     | 1     2     8     | 6     9     4     |
 |-------------------+-------------------+-------------------|
 | 349   2     579   | 35    8     47    | 1     6     39    |
 | 3489  1     579   | 6     59    47    | 2     58    39    |
 | 389   35    6     | 2     1     39    | 4     58    7     |
 *-----------------------------------------------------------*


The XY chain

9-(r2c1)-3-(r2c2)-5-(r9c2)-3-(r9c6)-9

eliminates the 9 in r9c1 since r9c1 sees both r2c1 and r9c6.

It's all singles from there.