The New Sudoku Players' Forum

[bruceka9]

Given this solution martix:


+---------------+--------------+-----------------+
| 5 19 7 | 4 3 6 | 19 8 2 |
| 19 4 6 | 2 8 5 | 7 19 3 |
| 8 3 2 | 7 1 9 | 56 4 56 |
+---------------+---------------+----------------+
| 4 6 5 | 1 9 7 | 2 3 8 |
| 7 8 3 | 6 4 2 | 159 19 15 |
| 19 2 19 | 8 5 3 | 46 7 46 |
+---------------+---------------+-----------------+
| 2 7 49 | 3 6 1 | 8 5 49 |
| 6 19 8 | 6 7 4 | 3 2 19 |
| 3 5 14 | 9 2 8 | 14 6 7 |
+----------------+--------------+----------------+

What is the logic to solve this? I can use guessing, but I think someone can walk me through the solution.

Thanks, Bruce

[emm]

bruce - you have 2 6s in row 8 - I guess r8c4 = 5

You can solve this by colouring the 1s. Label conjugate pairs alternately – say (+) and (-) and because they’re pairs, either the + or the - will be correct.

The rule is that if you find two of the same colour in the same group then they are the incorrect ones.

Code: Select all

+-----------------+-----------------+-----------------+
| 5     19+   7   |  4     3    6   | 19-    8    2   |
| 19-   4     6   |  2     8    5   | 7      19+  3   |
| 8     3     2   |  7     1    9   | 56     4    56  |
+-----------------+-----------------+----------------+
| 4     6     5   |  1     9    7   | 2      3    8   |
| 7     8     3   |  6     4    2   | 159    19-  15- |
| 19+   2     19- |  8     5    3   | 46     7    46  |
+-----------------+-----------------+-----------------+
| 2     7     49  |  3     6    1   | 8      5    49  |
| 6     19-   8   |  5     7    4   | 3      2    19+ |
| 3     5     14+ |  9     2    8   | 14-    6    7   |
+-----------------+-----------------+----------------+

You end up with two (-) both in r5 and in c7 => all the (-) are false
For a good explanation of colouring click here Simple Sudoku

The logic of chains is similar
r1c2=1 => r1c7=9 => r2c8=1 => r5c8=9
r1c2=9 => r8c2=1 => r8c9=9 => r5c9=1 => r5c8=9
=> r5c8 must = 9

Click here for SadMan's explanation of forcing chains

[Myth Jellies]

An easier way to solve it is using the BUG. You have only one three-candidate cell, and all the rest are two-candidate cells. To avoid the BUG you have to choose the selection in the three-candidate cell which appears three times in its row/column/box. Thus, r5c7 = 1 must solve the puzzle.

[bruceka9]

Sorry for the typo on row 8.

Thanks for the solution. I'll study the technique.

Bruce