The New Sudoku Players' Forum

[daj95376]

I can't find any instances of the pattern in your two examples, ...

That seems to be the common consensus from what I've read. However, everyone has done a lot of work on these two puzzles ... AND ... I'm going to carefully review all of the responses to learn what I can about the URs in these two puzzles.

Out of the 62 example puzzles posted by Mike Barker, there are a few others that I couldn't understand/resolve. I'll give everyone a break before presenting the next troublesome UR pattern.

Thanks to everyone for your effort !!!


Regards, Danny

_

[blue]

Now, it's time to see how much of the above logic holds together on Mike Barker's second puzzle for this pattern. Soggestions?

Code: Select all

 Puzzle #55: Mike Barker UR+3X/1SL
 +-----------------------+
 | 6 9 5 | . 2 . | . . . |
 | . . 3 | 4 . . | . . . |
 | . . . | 9 . . | . 1 . |
 |-------+-------+-------|
 | . . 9 | . . 2 | 4 . . |
 | . . . | . 3 . | 1 . . |
 | 4 . . | . . . | 3 5 . |
 |-------+-------+-------|
 | . 6 . | . . . | . . . |
 | . 1 . | 8 . . | . 6 . |
 | 9 8 4 | . 1 . | . . . |
 +-----------------------+

 +--------------------------------------------------------------+
 |  6     9     5     |  3     2     1     |  78    478   478   |
 |  1     27    3     |  4     678   578   |  2569 *29   *29+5  |
 |  27    4     8     |  9     67    57    |  256   1     3     |
 |--------------------+--------------------+--------------------|
 |  78    3     9     |  1     5     2     |  4     78    6     |
 |  28    5     6     |  7     3     4     |  1    *29+8 *29+8  | <- SL on <9>
 |  4     27    1     |  6     89    89    |  3     5     27    |
 |--------------------+--------------------+--------------------|
 |  35    6     27    |  25    479   379   |  2589  2489  1     |
 |  35    1     27    |  8     479   379   |  259   6     2459  |
 |  9     8     4     |  25    1     6     |  257   3     257   |
 +--------------------------------------------------------------+
 # 61 eliminations remain


I've marked a potential UR(29) at r25c89. For the UR+3X\1SL to crack the puzzle, we need to eliminate <2> in both of r5c89

Eliminating 2r5c9, leaves skyscraper <2>r35\c1, that eliminates 2r2c8 and cracks the puzzle.

Code: Select all

+------------+--------------+---------------------+
| 6   9   5  | 3   2    1   | 78    478    478    |
| 1   27  3  | 4   678  578 | 2569  (29)   (259)  |
| 27  4   8  | 9   67   57  | 256   1      3      |
+------------+--------------+---------------------+
| 78  3   9  | 1   5    2   | 4     78     6      |
| 28  5   6  | 7   3    4   | 1     28(9)  (89-2) |
| 4   27  1  | 6   89   89  | 3     5      (27)   |
+------------+--------------+---------------------+
| 35  6   27 | 25  479  379 | 2589  2489   1      |
| 35  1   27 | 8   479  379 | 259   6      2459   |
| 9   8   4  | 25  1    6   | 257   3      (257)  |
+------------+--------------+---------------------+

(2=57)r69c9 - 5r2c9   [ U = 57r69c9, Y = 5r2c9 ]

From that, we get: 2r5c9 => 9r2c9.
Then with the SL on <9> and the bivalue in r2c8, we have:

   2r5c9 => 9r5c8,2r2c8,9r2c9 ... UR contradiction => -2r5c9

[daj95376]

Code: Select all

--- UR+3X/1SL:   (reformatted)

 Constraint #1:
 includes the extra cell(s) "(ab)U..." such that "U" is a locked set which includes "Y",
 "abY" is seen by all of the cells of "(ab)U..." which contain elements of "Y",

 Constraint #2a:
 "(ab)U..." can contain "a", and
 "(ab)U..." can contain "b" if all of its cells which contain "b" are seen by "abX"
 => "b" can be removed from "abX". Similarly,

 Constraint #2b:
 "(ab)U..." can contain "b", and
 "(ab)U..." can contain "a" if all of its cells which contain "b" are seen by "abX"
 => "b" can be removed from "ab(Z)".

 ab     abX
         |
         |a
         |
abY     ab(Z)  (ab)U...


[Solution Withdrawn: matches case #2 in a posting earlier by blue.]

_

[blue]

To cover both of Mike's examples, I would suggest this alteration:

Code: Select all

--- UR+3X/1SL:   (re-interpreted)

 Includes the extra cell(s) "(ab)U" such that "U" is a locked set which includes "Y".
 The cells in "(ab)U" needn't be in the same line as "abY" and "ab(Z)" (as the diagram below might seem to suggest).

 Constraint #1:
 "abY" is seen by all of the cells of "(ab)U" which contain elements of "Y",

 Constraint #2a:
 Cells in "(ab)U" that contain "a", can see "ab(Z)" or "ab"
 Cells in "(ab)U" that contain "b", can see "abX"
 => "b" can be removed from "abX".

 Similarly,

 Constraint #2b:
 Cells in "(ab)U" that contain "b", can see "ab(Z)" or "ab"
 Cells in "(ab)U" that contain "a", can see "abX"
 => "b" can be removed from "ab(Z)".

 ab     abX
         |
         |a
         |
abY     ab(Z)  ... (ab)U

[daj95376]

To cover both of Mike's examples, I would suggest this alteration:

Code: Select all

--- UR+3X/1SL:   (re-interpreted)

 Includes the extra cell(s) "(ab)U" such that "U" is a locked set which includes "Y".
 The cells in "(ab)U" needn't be in the same line as "abY" and "ab(Z)" (as the diagram below might seem to suggest).

 Constraint #1:
 "abY" is seen by all of the cells of "(ab)U" which contain elements of "Y",

 Constraint #2a:
 Cells in "(ab)U" that contain "a", can see "ab(Z)" or "ab"
 Cells in "(ab)U" that contain "b", can see "abX"
 => "b" can be removed from "abX".

 Similarly,

 Constraint #2b:
 Cells in "(ab)U" that contain "b", can see "ab(Z)" or "ab"
 Cells in "(ab)U" that contain "a", can see "abX"
 => "b" can be removed from "ab(Z)".

 ab     abX
         |
         |a
         |
abY     ab(Z)  ... (ab)U

Mike Barker doesn't fully qualify several of his descriptions. For example, he never indications as to whether a pattern is in a band or a stack. He leaves it to the reader to consider both possibilities. As to (ab)U, I believe that he was only indicating that there must be at least one cell of (ab)U that sees abY ... and that Y must be limited to (ab)U cells that see abY.

Note: I an concerned that he marked Z as optional in ab(Z). This may be a typo. If so, maybe there are others.

_

[daj95376]

I can't find any instances of the pattern in your two examples, and I can't follow DPBs second chain, but after I search of the forum I found an example of a puzzle where the method works <here>.

It's just an AIC with a derived strong inference through the UR - not a separate technique.

Code: Select all

+--------------------------+--------------------------+--------------------------+
|   2     46        1      | 69      456     59       |   3       7       8      |
|   8     67abU   459      |267abZ  267abY     3      | 459       1     459      |
|   3     479     459      |78       478       1      | 459       6       2      |
+--------------------------+--------------------------+--------------------------+
|   6     489       3      |   1     258       7      | 4589    289     459      |
|   1       5       2      | 68        9       4      |   7     38      36       |
|   7     489     49       |   3     2568    25       |   1     289     4569     |
+--------------------------+--------------------------+--------------------------+
|   4       3       8      |279abX  27ab       6      | 29        5       1      |
|   9       1       6      |   5       3     28       | 28        4       7      |
|   5       2       7      |   4       1     89       |   6     389     39       |
+--------------------------+--------------------------+--------------------------+

a=2, b=7, U=6, X=9, Y=6, Z=6.

(7=6)r2c2 - (6)r2c45 = (9)r7c4[AUR27r27c45] - (2)r7c4 = (2)r2c4 => r2c4 <> 7.

[Withdrawn comment: I (yet again) incorrectly assigned the wrong cells to the SL.]

_

[sultan vinegar]

Nice try, but it doesn't match the pattern. Your referenced puzzle has three SLs in the UR: <2> in [r2], <2> in [c4], and <7> in [r7]. Your identify abY and abZ using the SL in [r2], and you used the SL in [c4] to derive the elimination in your chain.
_

Have another look. There is a SL on 2 in c4 between the abX and abZ cells (a is 2 in this example), just as in Mike's pattern. Yes, there happen to be other strong links in this particular UR but these are not used by the pattern. There is no SL between abY and abZ in Mike's pattern.

More generally,



5 Truths = {1V1 1C4 2N567}
6 Links = {789r2 12n4 2b2}
2 Eliminations --> r12c4<>2

As AICs:
(6)r1c4 = QNT(789)r2c3467[AUR12:r12c34] - (789=2)r2c5 => r1c4 <> 2.
(1)r2c4 = (1-6)r1c4 = QNT(789)r2c3467[AUR12:r12c34] - (789=2)r2c5 => r2c4 <> 2.

Like I said, nothing more than AICs with a strong link through the AUR. If you can do AICs with AURs, then you can do this. There are no new techniques to memorise.

[daj95376]

Sultan: I've updated my post above. Sorry over the confusion.

For the UR+3X\1SL to crack the puzzle, we need to eliminate <2> in both of r5c89.

My alternative is
*9r5c8->*2r2c8->2r3c1->*2r5c9->75r69c9->*9r2c9 => r5c8<>9

Here's the grid with the UR cells identified.

Code: Select all

  ab=r2c8;  abY=r2c9;  abX=r5c8;  ab(Z)=r5c9
 +--------------------------------------------------------------+
 |  6     9     5     |  3     2     1     |  78    478   478   |
 |  1     27    3     |  4     678   578   |  2569 *29   *29+5  |
 |  27    4     8     |  9     67    57    |  256   1     3     |
 |--------------------+--------------------+--------------------|
 |  78    3     9     |  1     5     2     |  4     78    6     |
 |  28    5     6     |  7     3     4     |  1    *29+8 *29+8  | <- SL on a=<9>
 |  4     27    1     |  6     89    89    |  3     5    U27    |
 |--------------------+--------------------+--------------------|
 |  35    6     27    |  25    479   379   |  2589  2489  1     |
 |  35    1     27    |  8     479   379   |  259   6     2459  |
 |  9     8     4     |  25    1     6     |  257   3    U257   |
 +--------------------------------------------------------------+
 # 61 eliminations remain


The UR+3X\1SL pattern only identifies possible eliminations for b=2 in cells abX(r5c8) and/or abZ(r5c9). Both eliminations are needed to crack the puzzle.

I've identified an (ab)U in r69c9 and derived:

Code: Select all

  2r5c9 -> r5c8=9, r6c9=7 -> r9c9=5 -> r2c9=9 -> r2c8=2 -> DP  =>  -2 r5c9