The New Sudoku Players' Forum

[balchsf]

Please help with this puzzle

XXX23X1X5
1XXX75XXX
5XXXX14XX
681927534
274583916
359146728
7X23XX851
X1XX52X97
9X5718X4X

Thanks,

[Carcul]

Code: Select all

 *--------------------------------------------------------------------*
 | 48     469    678    | 2      3      49     | 1      678    5      |
 | 1      23469  368    | 468    7      5      | 236    68     239    |
 | 5      2369   3678   | 68     69     1      | 4      678    239    |
 |----------------------+----------------------+----------------------|
 | 6      8      1      | 9      2      7      | 5      3      4      |
 | 2      7      4      | 5      8      3      | 9      1      6      |
 | 3      5      9      | 1      4      6      | 7      2      8      |
 |----------------------+----------------------+----------------------|
 | 7      46     2      | 3      69     49     | 8      5      1      |
 | 48     1      368    | 46     5      2      | 36     9      7      |
 | 9      36     5      | 7      1      8      | 236    4      23     |
 *--------------------------------------------------------------------*


[r8c4]-4-[r7c6]=4=[r7c2]-4-[r2c2]=4=[r2c4]-4-[r8c4]

which implies that r8c4 cannot be "4" and that solve the puzzle. Please note the following: in row 2, candidate "4" can only be in cells r2c2 or r2c4; in row 7, the same candidate "4" can only be in cells r7c2 or r7c6. Now, let's assume that r8c4 is 4: then r7c6 cannot be "4" and so r7c2=4, which means that r2c2 can no longer be "4" and so r2c4=4. But now we have two "4s" in column 4. So r8c4 cannot be "4".

Carcul

[ravel]

For me this was easier to spot (bivalue instead of bilocation):
r8c4=4 > r8c1=8 > r1c1=4 > r1c6=9 > r7c6=4

[QBasicMac]

Both of these require initial attention to r8c4=4:
[r8c4]-4-[r7c6]=4=[r7c2]-4-[r2c2]=4=[r2c4]-4-[r8c4]
r8c4=4 > r8c1=8 > r1c1=4 > r1c6=9 > r7c6=4

Something about that I like since it resembles my beloved T&E but is less work.

But I lack the intuition or rule that would focus me on that particular cell/value.

Is there any hint besides experience?

I would have just put 4 in the first cell to arrive at

Code: Select all

+-------------+---------------+--------------+
| 4  {9}  678 | 2    3    {9} | 1   678  5   |
| 1  239  68  | 4    7    5   | 26  68   239 |
| 5  239  678 | {8}  69   1   | 4   678  239 |
+-------------+---------------+--------------+
| 6  8    1   | 9    2    7   | 5   3    4   |
| 2  7    4   | 5    8    3   | 9   1    6   |
| 3  5    9   | 1    4    6   | 7   2    8   |
+-------------+---------------+--------------+
| 7  4    2   | 3    {9}  {9} | 8   5    1   |
| 8  1    3   | 6    5    2   |     9    7   |
| 9  6    5   | 7    1    8   | 23  4    23  |
+-------------+---------------+--------------+


(impossible)
and thus start over with 8, which solves easily.

But it was just the first random cell. I tried some others, such as r9c9.

Try 2 and only get to here

Code: Select all

+----------------+-------------+-------------+
| 48  469   678  | 2    3   49 | 1   678  5  |
| 1   3469  368  | 468  7   5  | 2   68   39 |
| 5   2     3678 | 68   69  1  | 4   678  39 |
+----------------+-------------+-------------+
| 6   8     1    | 9    2   7  | 5   3    4  |
| 2   7     4    | 5    8   3  | 9   1    6  |
| 3   5     9    | 1    4   6  | 7   2    8  |
+----------------+-------------+-------------+
| 7   46    2    | 3    69  49 | 8   5    1  |
| 48  1     368  | 46   5   2  | 36  9    7  |
| 9   36    5    | 7    1   8  | 36  4    2  |
+----------------+-------------+-------------+


Which requires nested T&E (for me).

So evidently, expertise in picking the attention cell is important.

Help?

Mac

[CathyW]

Colouring on 4s allows placement of 4 in three cells and complete solution follows quickly afterwards.

Code: Select all

 *--------------------------------------------------------------------*
 | 48a    469    678    | 2      3      49a    | 1      678    5      |
 | 1      23469a 368    | 468A   7      5      | 236    68     239    |
 | 5      2369   3678   | 68     69     1      | 4      678    239    |
 |----------------------+----------------------+----------------------|
 | 6      8      1      | 9      2      7      | 5      3      4      |
 | 2      7      4      | 5      8      3      | 9      1      6      |
 | 3      5      9      | 1      4      6      | 7      2      8      |
 |----------------------+----------------------+----------------------|
 | 7      46a    2      | 3      69     49A    | 8      5      1      |
 | 48A    1      368    | 46a    5      2      | 36     9      7      |
 | 9      36     5      | 7      1      8      | 236    4      23     |
 *--------------------------------------------------------------------*


Contradiction in box 1 and column 2 therefore all cells marked with A must be 4.

[ravel]

QBasicMac, i can only say that i make a quick look at each bivalue cell, if there is an xy-wing and follow some cells, that immediately follow from each of the 2 values. Thats, how i found this (and one a cell longer) in short time.
Two strong links (3 possibilities in CathyW's marked cells) would take me longer (without number filter).