is.gd/hoke545AnnualGoatX

Uh, this is tough. With a nice SI we solved hi/lo in A89 and now have 4 solved cells: A4, B45, G1. Anyone any chainfree ideas?

I didn't get further as well. No idea how to attack the diagonals.

An additional clue at E4 makes it solvable with simple standard tools, which is too easy for a jgrab.
So I am afraid you have to go through all the possible eliminations in the body of the goat, involving some simple and some more elaborated SI/head chains. But of course, you can find the value of E4 in the solution part of the extended URL

Finally done. Extra hard work with many SI/chains, (skewed) wings (as hoke mentioned) and only 1 Setti I have used to the end; a lot of logic to finding though.
One of the options in H2J1 collapses by using a chain where J3J7 = naked pair which gives H5… little chaining in columns 4, 5 and 6 leads to contradiction. Then the area around C7: there can be eliminated one number that defines hi/lo in row C. |D4 − E5| = 2. Right decision for C7. Position of 4 in row 1. Precise CE in E789. Right decision in F1. Chain on FGHJ6 leads to basic filling in.

Interesting puzzle. Thanks, hoke.

It's snowing where I live.

is.gd/morl545_1

Thanks for this puzzle. It was not that difficult. The final not too serious technical difficulties could be overcome by a combined setti on 1 and 9 which gave it a really nice touch.

Thank you, a very nice one. Not too difficult but not as easy as this week's "regular extreme".

22 min; very nice! At the end, a little 3-cell interaction replaces the combined Setti mentioned by AndreasM.

There's always aesthetic quality in your puzzles. I like that very much. Thank you!

really nice, ended it with a y-wing.

Qdennis:
y-wing - Congratulation, a new Str8ts-Rule is born.

A slightly easier X puzzle. No chains required.


is.gd/Leren_545_X2

Indeed, this one solves directly, not too easy, though. Thanks.

The Annual Goat
As in previous years, I gladly donated a goat for a worldvision project, expressing my thanks to all puzzle creators. Please receive your donation certificate from
is.gd/goat_for_creators
Another goat donation goes, as always, directly to Klaus, our Nestor of puzzle creation. Let's see whether he will again arrange for an Advent calendar ...

thanks a lot j,

as always, great stuff.

no calendar from myself this year, in times of mass consumption a little piece of chocolate a day seems outdated.

peace and love to all of you.

We'll hold your calendar actions in best remind. You're always right - stay healthy and have a good time!

is.gd/Leren_545_B2

one chain - very nice

Again chainless fun with CE and B Setti power. I'd like to point out two B setti moves I really enjoyed:

1. At an early point it's clear that 7 and 8 are missing from col 8. Therefore 7 or 8 must be missing from Box 3, therefore 7 or 8 must be missing from B, therefore 8 not in B34567.

2. After it becomes clear that 8 not in Box 3 it's clear that 7 not in Box 9 and therefore 7 not in H.

Here is another easy B Power move I came up with, which I call Box/Digit conflict.

eg. at 8 solved cells 9 in A1 would cause both A2 and B1 to be 2. Since this can't be True, A1 <> 9.

A 3 cell SI or maybe just call it a standard B Power move.

@ BP

wrt your 1. Early on you also know 7 is in B => - 8 in B and Box 3

@BP and Leren:
As I mentioned two weeks ago I'm new in this B-stuff. I can see that 7 and 8 are missing in col 8. So the "usual" rules say they must be missing in row B or H. It's easy to find 7 is missing in H (and must be in B).

But: How do I know that 7 or 8 must be missing in box 3??? (I assume 3 is ABC789, correct?)

@Berny: For B str8ts the following rules hold:
1. If a digit is missing from a box it must be missing from at least one of the rows and at least one of the cols intersecting with that box.
2. If a digit is missing from a row/col it mus be missing from at least one of the boxes intersecting with that row/col.

(The proof by contradiction is not difficult.)

In this case that means: Only 7 and 8 are missing from cols 7,8,9. Thus one of both digits is missing from box 3 the other is missing from box 9.

The above observation is really powerful in dealing with B str8ts. I use it all the time in different situations.

@BP:
Thank you, got it now! Indeed powerful...

is.gd/Leren_545_B1

is there a unique solution? I don't have one (F56=H56=56). Maybe I missed something!?

@ witch

In my solution there are no 5's in FH56.

thnx - now it's unique

Chainless fun, many thanks.

is.gd/Leren_545_X

Very, very hard. Took me 32 hours. All kind of tools, but mainly HCs.

Phew - took me longer than jgrab - really tough!
Loaded with dependencies, but couldn't find a chainless way, either. Thank you, Leren!

Sorry for the typo: should read "2 hours"

Seemingly easier than last week's X-puzzle as the high-low decisions are not that difficult to make. But solving it took me nevertheless quite some time. Thanks for the puzzle.

@Leren: Is there really no chainfree way forward? So far I got all hi/lo decisions and A79, D1, E3, F456.

@ BP

Look for a 3 cell SI in G3 & H23.

@Leren: Many thanks for the hint! Actually I didn't see an SI in G3H23, but staring at that area I noticed an x-power step I had previously overlooked. With that and two more short SIs the puzzle actually solved chainfree. A really great challenge, thank you for that!

@ BP

The SI was 5 r8c3 => 4 r8c2 => 3 r7c3 => - 5 r8c3.

Then 5's in H28 removes 5 from B28 etc

is.gd/Leren_545

After an uncharacteristically slow start it went rather smoothly. Thanks for this really fine puzzle.

BCA2 needed long chains, but solved.

Reserved for discussion of Andrew's puzzle.

Solvers job

Well, almost. You do one x-wing and one setti.

Like Jan, for me 1 hour of fun wc.

First I solved the puzzle without using a solver. It wasn't hard. I use two URs, one XW and two settis (on 1 and 8). Now I see that it could have been simpler to deal with. Thanks!