Hi guys,
I knew I could count on you 2 to give it a try.
Here is my solution:
Starting position:
Code: Select all
. . 3|. . .|. 6 .
. 6 4|. . .|9 . 8
2 . .|. . .|1 . .
-----+-----+-----
. 3 9|4 2 8|. . 6
. . .|. 9 .|. . .
4 . 2|. . 5|. . 9
-----+-----+-----
. . .|5 . 7|4 . 2
. . .|. . .|. . .
. . 8|2 . .|. 3 1
After basics,
a naked pair in column 7... more basics...
a naked pair in box 9,
the naked triple in column 1,
a hidden pair in row 8,
a naked pair in column 3... more basics...
brings us here:
Code: Select all
.------------------.------------------.------------------.
| 89 1789 3 | 189 14578 14 | 2 6 57 |
| 157 6 4 | 13 1357 2 | 9 57 8 |
| 2 789 57 | 89 578 6 | 1 4 3 |
:------------------+------------------+------------------:
| 17 3 9 | 4 2 8 | 57 157 6 |
| 68 5 16 | 7 9 13 | 38 2 4 |
| 4 178 2 | 6 13 5 | 38 17 9 |
:------------------+------------------+------------------:
| 369 19 16 | 5 38 7 | 4 89 2 |
| 39 2 57 | 138 1348 134 | 6 89 57 |
| 57 4 8 | 2 6 9 | 57 3 1 |
'------------------'------------------'------------------'
At this moment, you can find a swordfish for digit 5 in rows 1,3,8... which takes you on a path with more subsets, another swordfish, some coloring, an XYZ-wing culminating in a N-star constellation.
...but wait...there is an interesting development:
Code: Select all
.------------------.------------------.------------------.
| 89 1789 3 | 189 14578 14 | 2 6 57 |
| 157 6 4 | 13 1357 2 | 9 57 8 |
| 2 789 57 | 89 578 6 | 1 4 3 |
:------------------+------------------+------------------:
| 17 3 9 | 4 2 8 | 57 157 6 |
| 68 *5 16 |*7 9 13 | 38 2 4 |
| 4 *178 2 | 6 13 *5 | 38 17 9 |
:------------------+------------------+------------------:
| 369 19 16 |*5 38 *7 | 4 89 2 |
| 39 2 57 | 138 1348 134 | 6 89 57 |
| 57 4 8 | 2 6 9 | 57 3 1 |
'------------------'------------------'------------------'
Look at the cells I marked with a star. Normally, you do not care about cells that are solved, but take another look.
What if R6C2 contains a 7? Then there would be an isolated group of digits 5 and 7 in box 458, rows 567 and columns 246. This isolates the remaining candidates for 5 and 7 with no given value. As a result, there are 2 ways to place the remaining candidates 5 and 7. There would be at least 2 solutions to the puzzle.
Not when we assume a unique solution. So we can eliminate R6C2 digit 7 as a candidate, which allows us to solve the remainder of the puzzle with singles only.
This is a
reversed uniqueness situation, a.k.a. reversed BUG. I like it because it is so easy to see, it uses completed cells, and it is so easy to build into the next version of my solver
In this puzzle, it significantly shortens the solving path.
David, have you been solving the same puzzle?
Ruud.
