Though how difficult it is depends on your audience -- this is an easy puzzle for this forum -- I'd rate this one "medium" for the general public. This is the rating it would get in most newspapers. Solving without pencilmarks, using "crosshatching" only, there are two points in the process that would require something beyond the simplest logic. A skilled solver might not notice the difference, but a beginner might get stuck.
Most of the puzzle can be solved without pencilmarks with the easiest (and most fun) tactic, often called "crosshatching" (aka "hidden singles" or "pinned squares", etc).
For example, the 7s in rows 2 and 3 cross out all but one of the available cells in box 1 -- so you can place a 7 in row one, column 3.
- Code: Select all
+-------+-------+-------+
| 5 4 . | . . . | . . . |
| x 1 x | . . . | 7 . . | <--
| x x x | 7 . . | 8 6 . | <--
+-------+-------+-------+
| . . . | . . . | . 1 . |
| . . . | 8 . 3 | . . . |
| . . . | 2 . . | . 5 . |
+-------+-------+-------+
| . . . | . 1 . | . . . |
| . . 8 | . . . | 3 . . |
| 2 . . | . . 6 | . . . |
+-------+-------+-------+
That's all you need to do for most of the puzzle -- except two points:
The 3s in row 5 and row 8 cross out all but two of the cells in column 5, not quite enough to place a number:
- Code: Select all
+-------+-------+-------+
| 5 4 7 | 1 6 8 | . 3 . |
| 8 1 6 | . . . | 7 4 5 |
| . . . | 7 . . | 8 6 1 |
+-------+-------+-------+
| . . . | 6 x . | . 1 . |
| . . 1 | 8 x 3 | . . . |
| . . . | 2 x 1 | . 5 . |
+-------+-------+-------+
| . . . | . 1 . | 5 8 . |
| 1 . 8 | . x . | 3 . . |
| 2 . . | . 8 6 | 1 . . |
+-------+-------+-------+
Look in box 1 -- though there is no 3 placed there yet, you can see that it must be in row 3 -- so you can crosshatch one more cell from column 5 and place a 3 at r2c5. This is slightly more difficult than the simplest tactic, but I think most beginners could find this without needing pencilmarks.
A little later, the solver will come to this point:
- Code: Select all
+-------+-------+-------+
| 5 4 7 | 1 6 8 | 2 3 9 |
| 8 1 6 | 9 3 2 | 7 4 5 |
| . . . | 7 . . | 8 6 1 |
+-------+-------+-------+
| . . . | 6 . . | . 1 . |
| . . 1 | 8 . 3 | . 2 7 |
| . . . | 2 . 1 | . 5 . |
+-------+-------+-------+
| . x . | * 1 . | 5 8 2 |
| 1 x 8 | . 2 . | 3 . 6 |
| 2 x x | . 8 6 | 1 . 4 |
+-------+-------+-------+
There are three cells open in column 4 -- they must be 3, 4 and 5 in some order. Row 7 column 4 (with the asterisk) must be 3, 4 or 5. But there's a 5 in row 7 -- so it must be a 3 or 4. But the 4s in row 1 and 9 crossout all all but two cells in box 7, both of which are in row 7-- so you know that the 4 in box 7 must be in row 7 -- so now our target cell cannot be a 4 either, allowing a placement of 3.
This step is beyond most casual players to accomplish without at least a few pencilmarks, and this is what changes this puzzle from "easy" to "medium". Using pencilmarks, this step is no more difficult than the first sticking point.
