RW wrote:Nice puzzle, but unfortunately it is solved with an
 UR+2kx:
 

RW wrote:AS for difficulty comparison, I still think URs like this are easier than other patterns, because the pm grid always tells you exactly where to look for them. When you see a naked pair in the same unit, you know there's a potential UR, XYZ-wings don't reveal their existance quite that easily.
As I said above, in general I agree with you. I happen to think though that some of the UR deductions are a lot harder to spot than others (unless of course you're looking for them). And when you are looking for xyz-wings specifically, they are a breeze to find.
Perhaps this transported wxyz-wing will fare better. An assist goes to JPF for more or less
 finding this puzzle- Code: Select all
- . . .|9 . .|. . 8
 . 1 .|. 3 4|2 6 .
 . . .|8 . 2|3 . .
 -----+-----+-----
 9 . 7|. 2 .|. 3 6
 . . .|6 . 3|. . .
 6 2 .|. 9 .|5 . 1
 -----+-----+-----
 . . 5|2 . 7|. . .
 . 7 1|3 4 .|. 8 .
 3 . .|. . 6|. . .
 
After the easy stuff one gets to
- Code: Select all
- .---------------------.---------------------.---------------------.
 | 457    3      2     | 9      6      15    | 147    1457   8     |
 | 578    1      89    | 57A    3      4     | 2      6      79    |
 | 457    49     6     | 8      157    2     | 3      14579  479   |
 :---------------------+---------------------+---------------------:
 | 9      458-   7     | 145C   2      15D   | 48     3      6     |
 | 1      458    48    | 6      57-    3     | 4789   2479   2479  |
 | 6      2      3     | 47B    9      8     | 5      47     1     |
 :---------------------+---------------------+---------------------:
 | 48     6      5     | 2      18     7     | 149    149    3     |
 | 2      7      1     | 3      4      9     | 6      8      5     |
 | 3      489    489   | 15     158    6     | 147    1247   247   |
 '---------------------'---------------------'---------------------'
where a wxyz-wing implies that A, C or D is a 5. We may transport A to D via
[r2c6 = 5 = r1c6 - 5 - r2c4]
and hence either C or D is a 5. Thus, r5c5<>5 (also r4c2<>5), solving the puzzle.
An alternative way to make this deduction is the (2-color) nice loop:
r5c5 - 5 - {r9c5 = 5 = r9c4 = 1 = r4c4 - 1 - r4c6} - 5 - r5c5, => r5c5 <> 5.