I'll repeat the X link in a separate thread to make it easier to see.

is.gd/Leren_417_2X

Thanks Leren,

It occurs to me that with suduko only the major x/y diagonals are considered because all 9 digits are required, but str8ts is different. It should be possible to construct a puzzle where every diagonal must follow the same rules.

@ Ben

Your idea sounds over constrained to me. I tried it on a 3 x 3 Grid and could only get it to work (trivially) if r2c23 were all black, which effectively cancelled out the diagonals.

Thinking about the UR issue mentioned below, whilst AS is correct in saying that no Sudoku UR cell can occupy a SudokuX diagonal, the same cannot be said for all Str8ts UR's. It seems to me that a 4 cell UR is OK if any pair of its diagonally opposite cells is on the same diagonal, and if they are in the same diagonal compartment. In fact, a 4 cell UR could possibly have all 4 cells on both diagonals, as long as both diagonally opposite cells are in the same diagonal compartment.

Correction: I can get a 3 x 3 grid to look like :

123
3x2
23x

Seems to work but I can't see how you can fill in a diagonal cell r2c2 or r3c3 without a violation.

It's getting late in the week so I'll make these fairly gentle.

is.gd/Leren_417_2

is.gd/Leren_417_2X

Thanks for the x-version. Took me a while to see that the first diagonal is constrained right from the start. The rest, then, was standard work.

I am bored - any chance to get one more str8ts or Xstr8tsX ?

to be precise - for sure I would take also both :-)

Full support!

Support !!!

Yes, that'd be nice!

I would appreciate that too

What I really like about this weeks extreme is that a head-chain that leads to a solution can be applied right from the start, before solving any other cell. Consider the following:
- from A1, we can see that col 9 must exclude 1, therefore A9 must be 3,5,7,9
- from col 3, we can see that A2-4 must be low, A6-9 hi - this eliminates 3 from A9.
- if A9 is 5 or 9, we can see that A7=8, A8=6, A6=7, C6=8. But, if A9 is 5 or 9, C-F9 must be in range 5-9, and C89 would have to contain 8, which isn't possible.

Therefore, A9 must be 7. Having worked this out right at the start, the puzzle can then be solved easily.

Hi Ben,

yes thats nice.

for completeness sake i should be mentioned that the possibilities in H9 play an important role.
to start the chain with A6=7 -> C6=8 -> C9=3 -> A9=H9=9 seems a little more straight forward to me.

Hi KDN,

no result-Str8t this time? or only afterwards?

i always have trouble with this 'pure logic'-thing people use.

every valid Str8t is solvable with 'pure logic'. is there a 'non-pure' logic?
indeed, if you ran out of known elimination-strategies, and you try (some of) the finite possibilities, thats a very logical way to procede, maybe the only one.

i see it the same way
for me not solvable with "pure logic" one decision hi/lo Col 9 for me necessary

However, trial and error is not very elegant. Probably I did too much pure(!) mathematics in my life ;-) .

i think pure mathematics is full of trial and error. every proof by contradiction can be considered as such.

I completely agree with Klaus. Any simple elimination of a digit can be regarded as trial and error. And even the so called "brute force method" (trying out the candidates one after another) is based on pure logic!
I think what people mean by "pure" logic is: "obvious" logic, or, as kmr puts it, an "elegant" way to the solution. We feel better (smarter, or purer :)) whenever we discover a quick contradiction, instead of filling in too many fields (in mind or physically) without knowing the outcome.
In case of the current puzzle I doubt there is a way without a short chain (used same as Klaus, Ben and some others), of course based on pure logic. I find the chain very elegant, it can be done in mind and it makes the rest of the puzzle solvable by common tools.
In my view most of the puzzles without any chain are a nice pastime, but not a real challenge.

no brilliance needed. used something probably quite similar to Ben.

with column 9 you can smell the rat right from the beginning.

For once, this is an SI that I grok! 8-)

After that, the easiest way IMO is to use two 3-wings and a final single.

Oh, and for me the SI solves A9, not A7 as one comment has it. Dunno whether that's a typo or he used a different line of argument.

For me the SI was: A7=5 => A9=7 => D7=7 and HJ9=98 => 8 in every row but not in col 7. That's a contradiction so A7=8.

is.gd/Leren_417_Astrid_X

Ty Leren, i'm flattered <3 Both puzzles are very nice exercise.

Thanks. Nice demonstration of how the diagnonals work right from the start.

Thanks, Leren. Diagonals and settis solve it.

Hi Leren. Very nice puzzle - I really like your x-Variant ! UR-Arguments get quite different though! A well-used UR in regular str8ts can get unique by the additional constraints about the diagonals, so you have to be careful when using them. Keep on rolling with the Xs!

@ Andy

Andrew Stuart's advice for the X Sudoku variant is that UR's won't work if any of the cells is on a diagonal, otherwise they will work as normal.

@Leren
I'd say AS's advice doesn't apply in general. There are definitely UR-arguments that work, although they include diagonal cells. Depending on the particular nature of the argument, of course.

@ Astrid

I've been thinking that you might be right.

For example, if a 4 cell UR has two diagonal cells that are on the same diagonal then swapping the UR pattern may not produce a contradiction.

In the mean time I'm following AS's advice, which as least is safe.

I'm still busy developing my X solver by adding diagonal based moves. When I think it's reasonably complete I'll think more about the UR issue.

is.gd/Leren_417_Astrid

thanks Leren, very nice one!

Nice! Had difficulties to decide hi/lo for row A, however.

I used a UR-argument to avoid a UR in HJ 25, followed by settis and x-wings.