wapati wrote:This is pretty busy. Perhaps a finned sword, not sure if it is robust.
- Code: Select all
5 . 1|. 9 .|. . 2
. . 6|. . 2|. . .
2 7 .|5 . 1|. . .
-----+-----+-----
. . 5|7 . .|2 3 .
6 . .|. . .|. . 7
. 2 3|. . 9|4 . .
-----+-----+-----
. . .|2 . 6|. 7 1
. . .|9 . .|3 . .
9 . .|. 5 .|6 . 4
Simple stuff takes us to:
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.---------------.---------------.---------------.
| 5 348 1 | 468 9 38 | 7 46 2 |
| 348 3489 6 | 48 7 2 | 5 1 39 |
| 2 7 49 | 5 36 1 | 8 469 369 |
:---------------+---------------+---------------:
| 18 189 5 | 7 468 48 | 2 3 689 |
| 6 489 489 | 3 2 5 | 1 89 7 |
| 7 2 3 | 168 168 9 | 4 568 568 |
:---------------+---------------+---------------:
| 348 5 48 | 2 348 6 | 9 7 1 |
| 148 6 27 | 9 148 478 | 3 258 58 |
| 9 138 27 | 18 5 378 | 6 28 4 |
'---------------'---------------'---------------'
Note the potential deadly pattern in r12c24[48]. So r1c2=3, r1c4=6, r2c2=3 or r2c2=9.
If r1c6=3, then
r1c2<>3,
r2c2<>3 (via r9c6 and r9c2). Also, r3c5=6 and so
r1c4<>6. Moreover, we get r3c9=3 and so r2c9=9 implying
r2c2<>9. This contradiction shows that r1c6<>3 which solves the puzzle.
Notation guru's, please help me with the notation for this
My best guess is that it looks something like:
[r1c6](-3-[r1c2])(-3-[r9c6]=3=[r9c2]-3-[r2c2])-3-[r3c5]-6-[r1c4]=6|9=[r2c2]-9-[r2c9]-3-[r2c1]
and so r1c6<>3, solving the puzzle.