Variation of Locked Candidates

Advanced methods and approaches for solving Sudoku puzzles

Postby udosuk » Wed Jan 03, 2007 1:44 pm

ronk wrote:I agree, and think exemplars could be ...
Code: Select all
...

 "/" cells: neither X nor Y nor Z

 .  .  .  | XYZ+ XYZ+ /  | /  /  /    .    .  .  | XY ** ** | ** ** **
 XY ** ** | .    .    .  | .  .  .    XYZ+ /  /  | .  .  .  | .  .  .
 ** ** ** | .    .    .  | .  .  .    XYZ+ /  /  | .  .  .  | .  .  .
 ---------+--------------+--------    -----------+----------+--------
 .  .  .  | .    .    .  | .  .  .    .    .  .  | .  .  .  | .  .  .

For all four illustrations, both X and Y may be excluded from the ** cells, and the extra candidates (tagged '+') may be excluded from the XY+ and XYZ+ cells.

I think it must also be stated that Z is locked in those XYZ+ cells, otherwise they could simply be {XY} and the eliminations cannot be made...

Added later:
Come to think of it, there is another pair of "exemplars" you could make:
Code: Select all
 "/" cells: neither X nor Y nor Z

 .   .  .  | XYZ+ XYZ+ /  | /  /  /    .    .  .  | XYZ XYZ ** | ** ** **
 XYZ ** ** | .    .    .  | .  .  .    XYZ+ /  /  | .   .   .  | .  .  .
 XYZ ** ** | .    .    .  | .  .  .    XYZ+ /  /  | .   .   .  | .  .  .
 ----------+--------------+--------    -----------+------------+---------
 .   .  .  | .    .    .  | .  .  .    .    .  .  | .   .   .  | .  .  .

X,Y,Z may be excluded from the ** cells, and the + candidates may be excluded from the XYZ+ cells...:idea:
udosuk
 
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Postby ronk » Wed Jan 03, 2007 2:41 pm

udosuk wrote:I think it must also be stated that Z is locked in those XYZ+ cells, otherwise they could simply be {XY} and the eliminations cannot be made...

Agreed, but is it useful? The premise is no Z in the '/' cells, and no eliminations can be made in the row-box intersection AFAIK.

udosuk wrote:Come to think of it, there is another pair of "exemplars" you could make:
Code: Select all
 "/" cells: neither X nor Y nor Z

 .   .  .  | XYZ+ XYZ+ /  | /  /  /    .    .  .  | XYZ XYZ ** | ** ** **
 XYZ ** ** | .    .    .  | .  .  .    XYZ+ /  /  | .   .   .  | .  .  .
 XYZ ** ** | .    .    .  | .  .  .    XYZ+ /  /  | .   .   .  | .  .  .
 ----------+--------------+--------    -----------+------------+---------
 .   .  .  | .    .    .  | .  .  .    .    .  .  | .   .   .  | .  .  .

X,Y,Z may be excluded from the ** cells, and the + candidates may be excluded from the XYZ+ cells...:idea:

Right ... and keep on adding exemplar pairs, I suspect, until you run out of cells or have a smaller dual.

Conjecture: If you have a "naked ALS" in B\L (or L\B) -- where 'B\L' means the set of cells of a line subtracted from the set of cells of an intersecting box -- then all you need is a "hidden ALS" in L\B (or B\L) containing all the candidates of the "naked ALS." The only other requirement is that the size of the hidden ALS be greater than or equal to the size of the naked ALS. Lots of possibilites there.
ronk
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Postby Ruud » Thu Jan 04, 2007 12:29 am

I think that each Sue-De-Coq which only causes eliminations in one of the sectors has a complementary ALC move, but there is a selfdestructing variant of ALC which can also be used as a complement.

Here is an example:

Code: Select all
. . 6|9 . .|. . 4
. . 8|. . 4|. . 2
2 . .|. 3 .|. 8 .
-----+-----+-----
. 3 .|7 . .|. 9 .
. . .|. . .|. . .
. 9 .|. . 6|. 5 .
-----+-----+-----
. 6 .|. 4 .|. . 9
7 . .|1 . .|5 . .
1 . .|. . 2|7 . .

Here is a Sue-De-Coq:
Code: Select all
.------------.------------.------------.
| 3   5   6  | 9   2   8  | 1   7   4  |
| 9   1   8  | 5   7   4  | 36  36  2  |
| 2   47  47 | 6   3   1  | 9   8   5  |
:------------+------------+------------:
| 68  3   2  | 7   18  5  | 4   9   16 |
|#568#78 #157| 4   9   3  |*68  2   1-67|
| 4   9  *17 | 2   18  6  | 38  5   137|
:------------+------------+------------:
| 58  6   35 | 38  4   7  | 2   1   9  |
| 7   2   34 | 1   6   9  | 5   34  8  |
| 1   48  9  | 38  5   2  | 7   346 36 |
'------------'------------'------------'

This is the first ALC complement:
Code: Select all
.------------.------------.------------.
| 3   5   6  | 9   2   8  | 1   7   4  |
| 9   1   8  | 5   7   4  | 36  36  2  |
| 2   47  47 | 6   3   1  | 9   8   5  |
:------------+------------+------------:
|*68  3   2  | 7   18  5  | 4   9   16 |
|#568#78 #157| 4   9   3  |*68  2   1-67|
| 4   9   17 | 2   18  6  | 38  5   137|
:------------+------------+------------:
| 58  6   35 | 38  4   7  | 2   1   9  |
| 7   2   34 | 1   6   9  | 5   34  8  |
| 1   48  9  | 38  5   2  | 7   346 36 |
'------------'------------'------------'

This is the second (selfdestructing) ALC complement:
Code: Select all
.------------.------------.------------.
| 3   5   6  | 9   2   8  | 1   7   4  |
| 9   1   8  | 5   7   4  | 36  36  2  |
| 2   47  47 | 6   3   1  | 9   8   5  |
:------------+------------+------------:
| 68  3   2  | 7   18  5  | 4   9   16 |
|#568#78 #157| 4   9   3  | 68  2  *1-67|
| 4   9  *17 | 2   18  6  | 38  5   137|
:------------+------------+------------:
| 58  6   35 | 38  4   7  | 2   1   9  |
| 7   2   34 | 1   6   9  | 5   34  8  |
| 1   48  9  | 38  5   2  | 7   346 36 |
'------------'------------'------------'

Explanation:

r6c3 is the only cell in box 4 outside the intersection with row 5 that can contain 1 or 7. Because it has no other candidates, the intersection must contain either a 1 or a 7, but not both.
r5c9 is the only cell in row 5 outside the intersection with box 4 that can contain 1 or 7. Because the intersection cannot contain both of them, r5c9 must be 1 or 7, so we can eliminate the remaining candidate 6.

With 3 alternative views, the chance of spotting this elimination is greatly increased. For a huge Sue-De-Coq, there must be a smaller ALC complement.

Another note: These moves seem to be more abundant in the Sudoku-X variant. This may be the result of the selective eliminations caused by the diagonals.

Ruud
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Postby udosuk » Thu Jan 04, 2007 4:47 am

ronk wrote:
udosuk wrote:I think it must also be stated that Z is locked in those XYZ+ cells, otherwise they could simply be {XY} and the eliminations cannot be made...

Agreed, but is it useful? The premise is no Z in the '/' cells, and no eliminations can be made in the row-box intersection AFAIK.

The requirement that "Z is locked in the XYZ+ cells" is much more important than the premise "there is no Z in the '/' cells"... Why?

Consider the following situation:
Code: Select all
"/" cells: neither X nor Y nor Z

XYZ+ XYZ+ XYZ+ | XYZ+ XYZ+ / | / / /
XY   **   **   | .    .    . | . . .
**   **   **   | .    .    . | . . .
---------------+-------------+------
.    .    .    | .    .    . | . . .

Here, Z could be located in r1c123, and r1c45 can be {XY}, and we cannot eliminate X or Y from the ** cells...

On the other hand...

Code: Select all
"/" cells: neither X nor Y nor Z
"$" cells: neither X nor Y, but could be Z
"@" cells: anything but Z

XYZ+ XYZ+ XYZ+ | XYZ+ XYZ+ / | $ $ $
XY   **   **   | @    @    @ | . . .
**   **   **   | @    @    @ | . . .
---------------+-------------+------
.    .    .    | .    .    . | . . .

Here, the Z in b2 is locked in r1c45, so although r1c789 contain the candidate Z, we can still be sure r1c45 cannot be {XY}...
Which enables us to eliminate both X and Y from the ** cells...

And there is another extended case!
Code: Select all
"/" cells: neither X nor Y

.    .    .    | XY+ XY+ /  | / / /
XY   **   **   | ##  ##  ## | . . .
**   **   **   | XY  ##  ## | . . .
---------------+------------+------
.    .    .    | .   .   .  | . . .

Here, the case r1c45={XY} is again impossible, therefore we can eliminate both X and Y from the ** cells, as well as r1c6 and all the ## cells...:!:

:idea:
udosuk
 
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Joined: 17 July 2005

Postby ronk » Thu Jan 04, 2007 7:02 am

udosuk wrote:
ronk wrote:Agreed, but is it useful? The premise is no Z in the '/' cells, and no eliminations can be made in the row-box intersection AFAIK.

The requirement that "Z is locked in the XYZ+ cells" is much more important than the premise "there is no Z in the '/' cells"...

My statement was poorly phrased, as I certainly didn't mean that Z didn't need to be locked in the XYZ+ cells. I meant that, based strictly on one-box and one-line, no eliminations of Z were possible. IOW the exemplar was also representative of this ...
Code: Select all
 "/" cells: neither X nor Y nor Z

 .   .  .  | /  XYZ+ /  | /  XYZ+ /
 XYZ ** ** | .  .    .  | .  .    .
 XYZ ** ** | .  .    .  | .  .    .
-----------+------------+-----------
 .   .  .  | .  .    .  | .  .    .

... without any additional test to determine if the XYZ+ cells are restricted to a second box.

But that begs the question -- "locked how?" I can see only two possibilities: 1) relative to the entire row, or 2) relative to the row excluding the cells shared with the box, but neither possibility is making sense at this late hour.

Also, while I appreciate your effort in putting forward the two "extended cases", I'd like to see the discussion limited to deductions based on one-line and one-box -- at least for now.
ronk
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