Example for modified ALS xy wing rule

Advanced methods and approaches for solving Sudoku puzzles

Example for modified ALS xy wing rule

Postby bennys » Sat Dec 31, 2005 6:51 am

Code: Select all
 +-------+-------+-------+
 | 3 1 2 | . . 4 | 9 7 5 |
 | 6 . . | 1 9 7 | 3 2 8 |
 | 8 9 7 | . . . | . 1 . |
 +-------+-------+-------+
 | . . 9 | . 3 1 | 2 . . |
 | . . 3 | 2 . 8 | 1 . 9 |
 | . . 1 | . 4 . | . . 3 |
 +-------+-------+-------+
 | . 3 . | . 1 . | . 9 . |
 | 9 . 6 | 4 . 5 | . 3 1 |
 | 1 . 8 | . . . | 5 . 2 |
 +-------+-------+-------+
 
 +-------------------+-------------------+-------------------+
 | 3     1     2     | 68    68    4     | 9     7     5     |
 | 6    *45    45    | 1     9     7     | 3     2     8     |
 | 8     9     7     | 35   ^25    23    | 46    1     46    |
 +-------------------+-------------------+-------------------+
 | 457   45678 9     | 567   3     1     | 2     4568  467   |
 | 457  *4567  3     | 2    ^567   8     | 1     456   9     |
 | 257   25678 1     | 5679  4     69    | 67    568   3     |
 +-------------------+-------------------+-------------------+
 | 2457  3     45    | 678   1    %26    | 4678  9     467   |
 | 9    *27    6     | 4     278   5     | 78    3     1     |
 | 1    *47    8     | 39   %67    39    | 5     46    2     |
 +-------------------+-------------------+-------------------+

 
 
 A={R9C2,R8C2,R5C2,R2C2}
 B={R9C5,R7C6}
 C={R5C5,R3C5}
                                               
If A B ALS and C with freedom level of 2
x common to A,B,C  (2)
y restricted common to B,C (7)
z restricted common to A,C (6)

then  a cell that can 'see' all the x candidates of  A B and C can't be x.

and we get r8c5<>2 which solve the puzzle.
bennys
 
Posts: 156
Joined: Wed Sep 28, 2005 4:00 pm

Re: Example for modified ALS xy wing rule

Postby ronk » Sat Dec 31, 2005 7:39 pm

bennys wrote:
Code: Select all
 A={R9C2,R8C2,R5C2,R2C2}
 B={R9C5,R7C6}
 C={R5C5,R3C5}
                                               
If A B ALS and C with freedom level of 2
x common to A,B,C  (2)
y restricted common to B,C (7)
z restricted common to A,C (6)

then  a cell that can 'see' all the x candidates of  A B and C can't be x.

and we get r8c5<>2 which solve the puzzle.

Very nice. Here's another viewpoint for the almost locked sets ... the doubly-weakly-linked almost locked sets ...
Code: Select all
     other
       |
       A
     /   \
    z     x . . . x* (elimination)
    .     .     . .
    .     .   .   .
    .     . .     .
    z     x . . . x
     \\ //         \
       C            B -- other
     // \\         /
  other   y . . . y

Set properties:
   If A does not contain z, then A contains x
   If B does not contain y, then B contains x
   If C contains neither y nor z, then C contains x

For your example, the triple implication chains are:

r5c5=5 => r3c5<>5 => r3c5=2 => r8c5<>2 [edit: typo corrected]
r5c5=6 => r5c2<>6 => r8c2=2 => r8c5<>2
r5c5=7 => r9c5<>7 => r7c6=2 => r8c5<>2

P.S. I would normally assign 'z' to the elimination digit, and 'A' to the set with '2 degrees of freedom' ... but followed your labeling for continuity.
Last edited by ronk on Sat Dec 31, 2005 4:55 pm, edited 2 times in total.
ronk
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Postby bennys » Sat Dec 31, 2005 8:20 pm

You have typo in the first chain
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