nicktheconfused wrote:I'm stuck at this point. Can anyone help me?
Here are the original pencilmarks for your puzzle:
- Code: Select all
+------------------+-------------------+--------------------+
| 1236 7 139 | 169 5 4 | 8 129 129 |
| 12 4 159 | 7 189 28 | 3 6 1259 |
| 1236 13569 8 | 169 1369 236 | 2579 12579 4 |
+------------------+-------------------+--------------------+
| 7 38 4 | 589 389 1 | 259 2359 6 |
| 9 136 2 | 456 3467 3567 | 57 1357 8 |
| 5 1368 13 | 2 36789 3678 | 4 1379 1379 |
+------------------+-------------------+--------------------+
| 348 3589 359 | 4568 4678 5678 | 1 23579 23579 |
| 134 2 6 | 145 147 9 | 57 8 357 |
| 18 1589 7 | 3 2 58 | 6 4 59 |
+------------------+-------------------+--------------------+
You can proceed as follows:
> Locked candidate 3 in box 2
Translation: Look in the second box. Note that all 3's in that box are in row 3. That means that the 3 that you eventually place in box 2 MUST go on row 3. Thus all other 3's in row 3 can be erased. Erase 3 from r3c1 and r3c2 (r3c12).
> Locked candidate 6 in box 4
Same, except for column. Erase 6 from r3c2.
> Locked candidate 8 in box 4
> Locked candidate 1 in box 8
> Naked pair 57 in col 7
Translation: As r5c7 and r8c7 both contain 57, any 5 or 7 to be placed in col 7 MUST be in one of those two cells. Erase 5 and 7 from r3c7. Erase 5 from r4c7.
Here is your candidate list (also called "pencilmarks") at this point:
- Code: Select all
+-----------------+-------------------+------------------+
| 1236 7 139 | 169 5 4 | 8 129 129 |
| 12 4 159 | 7 189 28 | 3 6 1259 |
| 126 159 8 | 169 1369 236 | 29 12579 4 |
+-----------------+-------------------+------------------+
| 7 38 4 | 589 389 1 | 29 2359 6 |
| 9 136 2 | 456 3467 3567 | 57 1357 8 |
| 5 1368 13 | 2 36789 3678 | 4 1379 1379 |
+-----------------+-------------------+------------------+
| 348 359 359 | 4568 4678 5678 | 1 23579 23579 |
| 34 2 6 | 145 147 9 | 57 8 357 |
| 18 159 7 | 3 2 58 | 6 4 59 |
+-----------------+-------------------+------------------+
> Hidden single 7 in row 3
Translation: Note that r3 contains only one pencilmark 7 (in r3c8). Enter 7 as a solution there. That erases all other 7's in that box, row and column.
> Hidden single 5 in box 3
You now see that box 3 contains only 1 5. Place it and erase corresponding pencilmarks. Got it?
And so on. Here are some next moves...
Hidden single 5 in row 3
Hidden single 5 in col 3
Hidden single 5 in box 9
Hidden single 5 in row 9
r5c7 = 7
r9c9 = 9
r9c2 = 1
r9c1 = 8
Hidden single 1 in row 5
Hidden single 1 in row 6
Hidden single 9 in row 7
r2c3 = 9
Mac