Okay, found it, was an autofill issue in the new update.

Does anyone uses chrome? only happens in chrome for me, latest update.
Since 1-2 weeks, in all str8ts puzzles, when i click on an empty cell i get a few values to choose from a list instead of being empty. The list differ in values for each empty cell i click on. This list wasn't around before. Emptied the cache, images, even new install.
Anyone an idea?

is.gd/Leren_465_X

After 29 solved cells, when all 6,7 and 8 were clear, didn't find how to continue. Settis were mutually dependent. Probably a clever chain?

Arrived at the same position, also using a nice elimination on both diagonals with an isoceles triangle of cells.

A guess on C9 solves easily, but leading the other candidate to contradiction appears to be difficult. Can't see anything working with 1 and 9, either.

Look at the 9s in this position. In combination with Setti 3 you can exclude 9 in B9.

my key was F6, where the position of 7 was defined through the diagonal /. For the rest I used standard tools incl. one UR.
Thanks, Leren, nice puzzle.

Quite a challenge. Thank you, very nice.

@ABC: your hint did the job: it is then all about the mandatory 4 in B, 3 in col 9 and 3 in \ diagonal. Excellent!

At 29 solved, I considered the 1's. Specifically if H2=1, with cols 7&8, all 1's in col 6 are eliminated.

I don't follow ABC's hint.

Christoh _X_465
7 clues:
000000000
090000400
000070000
000000050
000000000
000003000
000000008
000000000
000008000
14 black cells:
001100000
110001000
000000100
000100010
100000010
000000000
100000100
010000000
000010000

TYPO:
correct: Christoph_X_465

is.gd/Christoph_X_465

Thanks, Leren, for the correct transformation.

Too difficult for me; mutual interdependencies on the diagonals. Borrowed the A1 hint from the headline, then it was easy.

Thank you Christoph (and Leren) for this nice puzzle. Solved it with CE, Settis without chains

@jgrab:
B9 was the key for me to clear the interdepency on the diagonals.

Puh...what a beast! Took me hours! Thank you Christoph for this challenging puzzle!!

A great puzzle, Christoph. Challenging, but very solveable.

First a simple SI solved the hi/lo in col 7. Then a short head chain to prove A2 couldn't be 7.

@Christoph:
If B9=5, J1=1; if B9=3, J1=3. E5=4 right from the start. So what does that help?
But now I see: If J1=1, a bigger chain leads - via B9 - to A1=J9=7, so J1=3.

@jgrab: dig. \ : 5 to be in C3 or J9. if B9=5 -> C3 = 5, thus 7 in H8 or J9, thus A1=68, thus A2 =7, thus C2 - G2 = 2-6, thus:
J1 = 1 and J2 = 1 , thus conflict. This solves.

@ Christoph: Thanks. Your proposal goes also across A1 and J9, like mine, but pursues a different idea. BTW: H8 can only be 68, after local elimination bottom right, so B9=5 immediately leads to J9=7.