The New Sudoku Players' Forum

[thierry marneffe]

Hello

I'm stopped by the following board (partially completed)

5.7861..2
289...176
.16927...
.7...8..4
..2.9.7..
6..27..1.
....42.8.
.23.89.5.
....1...9

Thanks for nay clue to continue solving this board ? What is logical technique to use ?

Thanks for your help

Thierry

[ravel]

Code: Select all

5.7|861|..2
289|...|176
.16|927|...
------------
.7.|..8|..4
..2|.9.|7..
6..|27.|.1.
------------
...|.42|.8.
.23|.89|.5.
....|1.|..9

Its a bit tricky:
Hidden pair in row 3: since you have 5 and 8 in box 1 and column 8, they must be in column 7 and 9 - no 3 there (especially not in r3c9).
Hidden pair in column 9: Since you have 17 in box 3 and 6, they must be in row 7 and 8 - no 3 there (r78c9).
Therefore the 3 in column 9 must be in box 6 in cells r56c9.
You can eliminate 3 from r5c8 (same box) and get a 6 there.

[daj95376]

Since you only posted the known values, it's impossible to tell which eliminations you've performed. Here's a worst case analysis on completing the puzzle. (2x) Naked Pair, Naked Triple, Locked Candidate (2). You get to find them!

[QBasicMac]

I assume you are trying to solve the puzzle without resorting to pencilmarks. Good luck!!

If you never heard of them, see this

Code: Select all

+-----------------+-----------------+------------------+
| 5     34    7   | 8      6   1    | 349    349   2   |
| 2     8     9   | 345    35  345  | 1      7     6   |
| 34    1     6   | 9      2   7    | 3458   34    358 |
+-----------------+-----------------+------------------+
| 139   7     15  | 1356   35  8    | 23569  2369  4   |
| 1348  345   2   | 13456  9   3456 | 7      36    358 |
| 6     3459  458 | 2      7   345  | 3589   1     358 |
+-----------------+-----------------+------------------+
| 179   569   15  | 3567   4   2    | 36     8     137 |
| 147   2     3   | 67     8   9    | 46     5     17  |
| 478   456   458 | 3567   1   356  | 2346   2346  9   |
+-----------------+-----------------+------------------+


It shows all possible candidates for each cell.

So you prepare a page like this and then do eliminations.

To start with, note that row 4 col 3 contains candidates 1 and 5. In other words, r4c3=15. But r7c3=15 too. So one of those cells MUST contain 1 and the other 5. Never mind you don't know which. The point is that you can eliminate candidates 1 and 5 from all other cells in col 3. There are only the 5's in r6c3 and r9c3, but every little bit helps. Erase those pencilmarks.

Similarly, 34 appears alone in two cells of row 3. Erase any other 3's or 4's in that row. Get the idea?

r7c9=137 and r8c9=17. 1 and 7 occur nowhere else in col 9. So you can erase the 3 to get

Code: Select all

+----------------+-----------------+------------------+
| 5     34    7  | 8      6   1    | 349    349   2   |
| 2     8     9  | 345    35  345  | 1      7     6   |
| 34    1     6  | 9      2   7    | 58     34    58  |
+----------------+-----------------+------------------+
| 139   7     15 | 1356   35  8    | 23569  2369  4   |
| 1348  345   2  | 13456  9   3456 | 7      36    358 |
| 6     3459  48 | 2      7   345  | 3589   1     358 |
+----------------+-----------------+------------------+
| 179   569   15 | 3567   4   2    | 36     8     17  |
| 147   2     3  | 67     8   9    | 46     5     17  |
| 478   456   48 | 3567   1   356  | 2346   2346  9   |
+----------------+-----------------+------------------+


Now here is an interesting case. Note that 3 appears only twice in column 9. One of those two cells MUST be 3. So all those other cells in the box cannot have 3's!. When you erase them all, you get r5c8=6. A cell is solved! Do that and at the same time erase the other 6's from the box, column and row.

Hope that helps.

Mac

[thierry marneffe]

Thanks for your replies ..... I actually use pencil marks (I guess it's impossible to do without) but solutions didn't come out ....

Thierry