The New Sudoku Players' Forum

[Carcul]

I have already described two examples of almost unique patterns: almost unique rectangles and an almost unique loop. In this post I will illustrate a slightly different almost unique pattern. Consider the following puzzle (“unlimited088” from Angus)

Code: Select all

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


and let’s jump to here:

Code: Select all

 38   7  5 | 2 6  1 | 38  4   9     
 49   2  38| 5 48 39| 1   7   6     
 1    49 6 | 7 48 39| 2   358 58   
-----------+--------+------------
 238  1  38| 9 5  28| 7   6   4     
 2689 69 7 | 4 3  28| 589 58  1     
 89   5  4 | 1 7  6 | 389 38  2     
-----------+--------+------------
 7    8  2 | 6 1  5 | 4   9   3     
 46   46 1 | 3 9  7 | 58  2   58   
 5    3  9 | 8 2  4 | 6   1   7     


If we look carefully to this grid, it can be concluded that we cannot have r5c7<>9 and r3c8<>3, or we would be left with a Swordfish pattern of “58s” that imply more than one solution to the puzzle. So, if “9” is not in r5c7, “3” must be in r3c8 and vice-versa. This can be expressed by the link [r5c7]=9|3=[r3c8], and we have

[r5c7]=9|3=[r3c8]-3-[r1c7](-8-[r5c7])-8-[r8c7]-5-[r5c7], => r5c7=9

which solve the puzzle.

Carcul

[Carcul]

[Post deleted]

[Jeff]

Hi Carcul, I am not really good at uniqueness patterns and please correct me if I am wrong. I just would like to understand whether it is okay to have a uniqueness pattern with the node [38] appears only once in a column 4 and column 6.

[Carcul]

Hi Jeff.

You are right: only if in box 2 both nodes "38" were in column 4 or 6 we would then have an AUP. In general we have the following:

We have an Unique Pattern on two candidates "x" and "y", if we have 2n cells populated only by "x" and "y", and distributed by exactly n boxes, n rows, and n columns (where n>1).

(I guess this "rule" can also be applied in the Extended Uniqueness Theory of Myth Jellies). In my grid, we would have eight cells with "3,8" distributed by four boxes, four rows, but five columns. Thanks for your observation Jeff, and my apologies to all users of this forum. However, after examining more carefully the grid I have noted that the same deduction can be made in a loop with the aid of a "new" extension of an existing technique that I have "discovered" some days ago, and about which I am writing a thread.

Regards, Carcul

[JeffInCA]

Carcul,

I've "read" some of the uniqueness posts here, but never really digested the entire concept.

This rule you posted is excellent. It really helped me visualize the foundation of the theory very quickly. Now maybe I'll be able to spot these scenarios more easily in puzzles.

Just one question:

Don't you need to add "where n > 1" to your rule? When n=1 I believe you simply have a naked pair.

Cheers,

Jeff

[Myth Jellies]

To help visualize Carcul's BUG-Lite pattern...

Code: Select all

 38   7  5 | 2 6  1 | 38    4     9     
 49   2  38| 5 48 39| 1     7     6     
 1    49 6 | 7 48 39| 2    *58+3 *58   
-----------+--------+----------------
 238  1  38| 9 5  28| 7     6     4     
 2689 69 7 | 4 3  28|*58+9 *58    1     
 89   5  4 | 1 7  6 | 389   38    2     
-----------+--------+----------------
 7    8  2 | 6 1  5 | 4     9     3     
 46   46 1 | 3 9  7 |*58    2    *58   
 5    3  9 | 8 2  4 | 6     1     7     


Just for grins and giggles, here is another BUG-Lite(+4) pattern. Probably too many extra candidates to be of much use, but they are as much fun to find as BUGs

Code: Select all

*38      7     5  | 2  6     1  |*38     4      9     
*49      2    *38 | 5 *48   *39 | 1      7      6     
 1      *49    6  | 7 *48   *39 | 2     *38+5   58   
------------------+-------------+------------------
*38+2    1    *38 | 9  5     28 | 7      6      4     
*69+28  *69    7  | 4  3     28 | 589    58     1     
 89      5     4  | 1  7     6  |*38+9  *38     2     
------------------+-------------+------------------
 7       8     2  | 6  1     5  | 4      9      3     
*46     *46    1  | 3  9     7  | 58     2      58   
 5       3     9  | 8  2     4  | 6      1      7     


[Carcul]

Hi JeffInCA.

Don't you need to add "where n > 1" to your rule? When n=1 I believe you simply have a naked pair.

Yes, I didn't state that because its obvious. But I will add that constraint to the above "rule". Thanks.

Regards, Carcul

[Carcul]

Here is another good example. Consider the following puzzle (which was posted by Havard in the thread "Is there a simple alternative for this template discovery?", page 4),

Code: Select all

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


which after the basic steps becomes

Code: Select all

 5    3    2  | 9    467  1    | 8    67   467 
 1    8    4  | 2    367  67   | 5    9    367 
 7    9    6  | 5    348  48   | 1    2    34   
--------------+----------------+----------------
 9    46   1  | 468  5    678  | 2    3    678 
 2    46   5  | 1    678  3    | 467  678  9   
 3    7    8  | 46   9    2    | 46   5    1   
--------------+----------------+----------------
 468  2    9  | 3    468  5    | 67   1    678 
 68   1    7  | 68   2    9    | 3    4    5   
 468  5    3  | 7    1    468  | 9    68   2   


Here we have an Almost Unique Pattern in cells r4c2/r5c2/r4c4/r6c4/r5c7/r6c7, and we can write

[r4c4]=8|7=[r5c7](-7-[r5c5]=7=[r4c6]-7-[r2c6]-6-[r9c6])-7-[r7c7]-6-[r7c5]=6=[r8c4]=8=[r4c4]

which implies r4c4=8 and that solve the puzzle.

Carcul