SudoCue Users

how do i post the grid with my possibles?

Assuming you're using Sudo Cue,

press {Control}+{M} or choose the menu option Edit/Copy Marks.

Paste the grid here in your topic, then put Code tags around it.

Good luck,

Ruud.

.---------------.---------------.---------------.
| 35 358 358 | 4 7 1 | 6 2 9 |
| 9 4 6 | 3 58 2 | 178 57 15 |
| 7 2 1 | 68 9 568 | 38 4 35 |
:---------------+---------------+---------------:
| 356 3578 358 | 678 1 9 | 2 57 4 |
| 56 9 2 | 67 3 4 | 17 8 156 |
| 1 378 4 | 5 2 68 | 37 9 36 |
:---------------+---------------+---------------:
| 4 1 35 | 2 58 358 | 9 6 7 |
| 2 6 9 | 1 4 7 | 5 3 8 |
| 8 35 7 | 9 6 35 | 4 1 2 |
'---------------'---------------'---------------'

hi garug,

welcome to this forum.

From the candidates you posted, I can only conclude that you have a very difficult Sudoku at hand.

The first step is not so complicated. There are locked candidates for digit 3 in row 4, box 4.

After that you need advanced colouring and tabling.

This puzzle may be a little to difficult for a beginning player. Start practicing with a few easier puzzles and move on to more difficult ones when you have mastered advanced solving techniques. Check the opened and generated puzzles with F7, to see what solving techniques are required.

Ruud.

I am not sure what that means.

Could it mean that the 3, 5 & 8 are the only candidates for Row 4, box 4?
That would place a 6 & 7 in that box....& maybe solve the puzzle!!

Thanks for your help, Rudd.

Sorry, it is not that simple...

The first step would be eliminate candidate 3 from R6C2, leaving only 7 and 8.

The second step eliminates 5 from R4C2 through colouring.

The next step tries candidate 5 in R1C1, which fails, so it can be eliminated.

The next step tries candidate 5 in R2C9, which fails, so it can be eliminated.

The next step tries candidate 5 in R3C6, which fails, so it can be eliminated.

Only then the puzzle is broken.

Ruud.

If you prefer patterns to chains, you might use a four star constellation after Ruud’s first step. The alpha star, cell (6, 9) containing (356), requires 3 to be eliminated from cell (3, 9). It therefore contains 5 and the rest is elementary.

Steve

Ooooh. New term. Can you define it, or point us to a URL that does
so?

Thanks very much,
Laura

The constellations were observed from Munich by someone_somewhere. They include four-star, five-star and other varieties. The constellations are interesting and reasonably common in difficult puzzles but not easily identified – my computer found this one when testing my beginner’s solver program.

Steve