Hi Andrew.

I don't know if you are reading this. Do you think it would be feasible to add a simple way to say "thank you" in the comment structure?
I would often just like to leave a "like" to the many creators without writing a long text, which disturbs the reading flow of the different solution strategies, especially with more complicated puzzles.

Otherwise, a big thank you for your work!

Full ACK (I feel the same - I hope, the many creators are roughly aware of how grateful I am for their puzzles <:-)

is.gd/ben_621_X

Firstly looked almost unsolvable, eventually not so hard. Either several SIs / short chains or one mid-long (which results to no 5 on the A9J1 diagonal) + one skewed X-wing + one Setti are enough.
Thanks…

There must always be a 5 on both diagonals??

Indeed, looks problematic at first glance, but then solves nicely. Some logical interdependencies clear the range of both middle diagonal strings which solves the corner points of col. 9. From there on, everything proceeds smoothly.

Coming back to hone's Bellona is.gd/hone620_Bellona puzzle from last week. Fantastic as always, thank you very much, hone. Like Ben, I avoided using obvious URs. Takes quite some time, though, only standard advanced techniques are required: setti, (not even very large) fishes, singles

To the str8ts-experts: double setti like 45 allowed or not? Thanks for an answer in advance.

Yes marianne, combination setti logic certainly can be used in some puzzles

v.gd/kst_621_BX

Very smooth and str8forward. Thank you!

Yes, very nice and chainfree. Thanks!

Perfect, thank you.

Chainfree but some careful looking is needee. Thanks!

v.gd/kst_621_B

Real hard CE work - still no chain required

v.gd/kst_621_X

Finally solved, yet, I gave up finding a chainless solution path

Solved it by using basic X-power, SIs (one of them more complicated, but a Setti consideration ‘saved’ it to call the process as a no-chain), a naked quadruple, an X-cycle, a skewed X-wing, and multi-Setti consideration to finish with singles.
After how many solved cells did you stick, Toni? Can help…

Thanks kozibrada for offering support. Well, I have reached a situation with the follwoing cells solved: A1678, B4567, C15678, D6, E5, F26, G7, H8, J89. On the way towards this situation I already made use of an "SI" that allowed to exclude 6 from A1 (any position of 4 in the lower right diagonal leads to that conclusion) - I guess some users would already consider this to be a (head) chain. But in the described situation my way forward was to exclude 4 from F9 (which yields G9=4). Though, the setti based contradiction is not difficult to find, this definitely goes beyond an "SI" for me. Once we have deduced G9=4 the puzzle basically collapsed (though I still needed yet another small SI to exclude 8 from D3. I would be interested your solution path after having reached the situation described above before I excluded 4 from F9 with a chain. Thanks in advance!

Maybe a small add-on to my explanations. The contradiction for my chain to exclude 4 from F9 ends with a setti 2 contradiction (still doable as a head chain, but too many small conclusions to consider it to be an "SI" from my point of view)

Ok, the situation with those 21 filled cells is right there I supposed. Still I went other way – the only two SIs I used have been related to 1 in H8 or J9 (J9 is a simple SI; H8 that more complicated, but not so extra: Setti on 4 + J89 = 23 produce a quick conflict with J4). Then we had probably the same thinking…

1) Instead of 4 in FG9, I noticed X-cycle-like logic in ABC234 for 9, which is actually a pair [89]: the cycle has five, odd number of members, which is wrong, thus AC4 ≠ 9. 2) This results to the skewed X-wing B2C3XBC9 → col. 9 ≠ 1. 3) Back to col. 4: full Setti on 6 means that row D contains surely 67 → full Setti on 7 → col. 2 ≠ 1. 4) CE in row D → solved H3459, G2, and also D5 (Setti). 5) Now the crucial multi-Setti consideration: row D cannot contain all 2, 5, and 8… but 5 in E4, and 9 in A23 just force into this situation. So E4 ≠ 5, A23 ≠ 9 → easy solving.

Thanks for sharing your findings. The X-cycle is something I missed for sure. I can also follow all the rest of your explanations. The multi setti argument on E4 is brillant! With regards to the argument on A23 - here it is even a bit more tricky to see the setti 2/5/8 conflict. Chapeau for this fantastic solution path! You convinced me that one can consider the puzzle to be solvable without chains :)

Ugh, I'm stuck much ealier 8-| - in fact, after only the unavoidable D6 has been solved. There is a simple SI/head chain eliminating 4 from E7 with the help of the diagonal, but that doesn't bring any progress.
What am I missing...?

@Jan: As written above - you can consider both remaning alternatives of 4 in the lower right diagonal. Both will lead you to the conclusion that 6 must be part of the lower right diagonal which again leads to A1=7. One of these options gives this conclusion directly because of a resulting naked pair 56, the other option is a bit more tricky. It includes a setti consideration that 4 would have to missing in col D, which would lead to an elimination of all possible options for 1 in the lower right diagonal. Hope that helps.

Sorry, of course, I mean col 4 not col D.

Thanks, Toni, for the appreciation. To ‘beat a beast’ with open eyes is always a great feeling. :-)
Last words to the puzzle: the A23 consideration about 9 within the 2/5/8 isn’t so pure (a slight SI) as 5 in A4 is eliminated…

@ Jan

E5 can be narrowed to two candidates because of the implications of B7 on the diagonal. That should help you narrow some choices...

v.gd/kst_621

Some nice 30 mins, thanks. Settis, CE, and a 3-fish towards the end.

My solution: one of the three possible numbers in J1 leads to the same numer in H2 and J2

Discussion of Andrew's Extreme Str8ts Puzzle, Weekly Edition #621, May 15 - May 21:

a UC helped to carry on after 12 solved cells

What means UC ?

UC=UR. Sometimes URs have no rectangle,so unique cell would be a better name.

not easy for me,
if you know that column 1 must not contain the number of Olympic rings, you can solve with normal tools

@ Augenweide: where did you see the UR/UC

Perhaps I missed something, because after the good hint from Augenweide I needed a chain.
Because of the UR/UC in A2, there is at least a 4 in A59 or EJ2

If there is no 4 in EJ2 => 4 in A59 => no 4 in row B and D (seti) => 8 in row B and 2 in row D => 2 and 8 in col 4, which is impossible.

This means H2=4 and among others „column 1 must not contain the number of Olympic rings“ (the hint from hp)

@MME: Thank you, and this is definitely a chain I should have described properly.
@grusto: sorry for being superficial.
A nice cooperation of UC and BCA, but far from obvious

Can someone explain UC in A2 in more detail. I have A2 = 1 or 4 and possible 4 in ABD and 2 and no 4 in 1 so there must be at least one 4 in rows ABD.

For me this puzzle was also too hard, but the solution of MME was perfect level of difficulty ;-) The most difficult step was:
If there is no 4 in EJ2 => 4 in A59 => no 4 in row B and D (seti)

In more detail. No 4 in EJ2 has two consequences. First, the UC says, that there must be a 4 in A59, otherwise we have 2 solutions for this puzzle, 4 and 1 in A2.
Second there is the Setti rule, which needs a missing 4 in a row due to the missing 4 in column 2. Together with the missing 4 in column 1 we now have both remaining rows B and D without 4.

This leads to a conflict, so the assumption "no 4 in column 2" must be wrong.

Like Greg, i don't understand what is the UC/UR in A2.
Please someone can explain it?

The U stands for uniqueness. A 4 in A2 would offer room for a second solution to this puzzle. Advanced strategies are explained in Slowthinkers slides.

Hello hoke, I followed most of your explanation but not all.

I see a 4 as possible in H2 so when you say "No 4 in EJ2 ..." then I understand you to be saying "Assume there is no 4 in EJ2 ...". So that is a fine way to start.

Then every logic after that is good. Assume a 4 in A59 to make A2 unique means A2 must be a 1 and therefore there are no 4's in column 2. Already there is no 4 in column 1 so rows B and D must have no 4. That satisfies the Setti rule.

Then you say "This leads to a conflict" but you don't say what the conflict is.

And further, you say 'the assumption of "no 4 in column 2" must be wrong' but the original assumption was "No 4 in EJ2" which is a different assumption.

Greg, you are right. The wrong assumption is "no 4 in EJ2" with the result that MME already mentioned. And also the conflict:

=> 8 in row B and 2 in row D => 2 and 8 in col 4, which is impossible

After A2 – UC, there is another possibility to break the puzzle alias underestimation of Setti on 9:
If F1 = 5 → H2 = 5 ≠ 4. We have surely four, maybe five 9’s in cols. This means that at least one of rows A and B must contain 9. But neither A5–9 ≠ 9 (due to the UC; E–J2 doesn’t contain 4) nor B6 ≠ 9 (full Setti on 5). Therefore F1 ≠ 5, then Setti on 5 solves the rest.

Thanks hoke. I don't think I've focused on UCs in that way before so that is something to add. Usually when someone says UR or UC then I'm looking for direct elimination and not something based on a chain of other issues so I'd call that a UC setti chain but since you can do it in your head then doesn't count as a "cheat".