is.gd/morl456

With 14 cells solved after settis and wings, there's an obvious UR, but it doesn't seem to help much. All I can do is a long contradiction chain on a cell with two candidates; the other solves easily.

Hmm, after 14 cells solved, I found an 'obvious' UR that helped a great deal, solving the entire top-left corner. Another UR removed a single candidate. Then a short SI solved G8 and I was done.

A good puzzle with a good level of complexity.

The puzzle is solvable w/o the UR with a single chain (lengh: 5 cells) that consists of only 2-candidate cells.

Typo correction: length: 5 cells...

is.gd/hoke456X

smooth X-puzzle with a nice interaction of the diagonals
thx hoke!

Nice and relaxing, many thanks!

Thanks. Nice; needed a short head chain to clear A9 after I got stuck.

Like Sonja, nice and relaxing puzzle, wthout Si or chains.

Not so easy, took me a while (BCA on 2) to find the clue.
Thank you, hoke!

At ten cells solved, one is tempted to use the UR at J9. But this is wrong, because that cell is coupled via A1 and C3 to the rest of the board. But it does give rise to a short, five-cell chain that solves G9 and with it the puzzle.

Maybe a stupid question: A1 and J9 - are they single compartments or do they belong to the diagonal?

@Ollie: Belong to the diagonal

@Others: Used like jgrab a very short SI/HC on A9 after 10 cells.

Could anyone explain how to solve completely WITHOUT a SI???

ok ... really silly question ... of course one big compartment

After the basics, the \ diagonal has its only 9s in A1 and J9. If A1=9, A5<>9 so G5=9, and if J9=9, G9<>9 and again G5=9.

@Morl:
yep... I guess that way is similar to the A9 approach (4 in A9 forces A5 and J9 to be 9, so no 9 left in G), but isn't that a SI as well?

My favorite SI was, that 2 in E7 would remove all 2s in the \ diagonal. But yours is even more str8tforward.

Just like hoke, it was E7 to me. BCA for 2 says: 2 must be missing in E or F; with E7=2 that would not be possible, because \ requires F6=2.
However, morl's and Berny's way ist more to the point, them medal goes to them!

Basically like Berny & morl, I used the 9's. My conclusion was that either 9 in the diagonal set G9=7. So a 3rd way of seeing the same thing.

I this hoke and Augenweide are seeing differently also. Hoke talks about an SI, so I read that as E7=2, E5=4, F5=25 - removing all 2's from diagonal.

But Augenweide's description is pure BCA 2 is missing from with row E or F, but if E7=2, and BCA removes 2 from F6, ... So it's a non-SI solution in my mind.

My (lucky) failure was that I had simply overlooked the potential candidate 9 in J9. Thus it was straight forward.

@Christoph: Ah ok, thanks for the info! That explains the situation...

As I mentioned solving G9 above, you can deduce I used Ben's argument.

is.gd/hone456_Rain

Thanks. 36 min; develops nicely after CE and some Setti calculation on B1 etc.

2B or not to be, this is the crucial cell.
Very nice, thank you, hone (and Shakespeare)

With 13 solved cells, I need a short SI in columns 6/7.

Thank You hone, solved it with an 8-fish and UR. No chain, no SI.

Very enjoyable, but we needed Augenweide's (and Hamlet's) hint to avoid chains. Thank you very much!

I agree, you need either the UR or the SI, both of which lead to the same conclusion about B2.

wow, great, finally here it is, good luck

I was glad to see new puzzle. Thanks. After CE and some x wings I found a naked triple in row B, which solves all puzzle.