No chains, SI's or UR's.


is.gd/Leren_500_X_M

Indeed, no complications. Thanks.

Can anyone explain who x-puzzles work? I have no idea what this means and cannot find a clue anywhere.

Thanks, Leren, a really nice exercise and a welcome break from staring at Leren_500_X which I have been doing for two days now always on the lookout for the non chain that will bring me to the 29th solved cell!

@Gue: In X-puzzles the str8ts rules also apply to the diagonals not only to the rows and columns. In this puzzle therefore there is a compartemnet of length 9 from A1 to J9, a compartment of length 6 from F4 to A9 and one of length 2 in J1H2. These compartments like those in rows/columns have to be filled with sets of numbers with no gaps and in any order. For example in this puzzle because of A9=3 it's immediately clear that J1H2 is either 12/21 or 78/87 or 89/98.

@Gue: the X-puzzles work exactly the same as the normal one with the addition that the diagonals behave like the rows and columns, i.e. each compartment in the diagonals also must contain a straight (if there are no black cells dividing them in compartments, then the whole diagonal must contain a straight). It's an X-Puzzle because the two diagonals form an X.

Sorry, hadn't noticed BP already answered Gue's question

Thank you for the explanation of X-puzzles. Now I can try to solve them... ;-)

In other words: An X-puzzle is a special sort of normal str8ts and so not every str8ts is an X-puzzle. Right?

@ Gue

It's a convention to end the link to the puzzle with _X to let you know that it's an X Puzzle.

Also, if you check out the list of X moves in the other thread you can use it as a primer for the many moves that are peculiar to X puzzles.

This develops nicely, with a single coming to the rescue when nothing else seems to work, until there is a position with only (567) and two 9s remaining. At that point I used an equilateral triangle on both diagonals and containing E5 to eliminate one candidate, which solves.

is.gd/hone500_Key

Thanks. Hard work; but step by step doable. A 4-fish decides A3, and two little head chains in the upper right corner.

Well, I'd say that 4-wing solves A3 somewhat indirectly...and only one head chain is required to eliminate a candidate from A9, with a setti as a chaser the rest is quickly done.

Thank you, hone, very nice and relaxing. I think I did like Jan. A little SI on A9, then a certain setti consideration opens up the puzzle. So strictly speaking, no head chain was neccessary.

is.gd/Leren_500_X

No idea how to come to a solution wthout guessing. So I give up.

I think 19 cells can be solved by standard (x-)tools. After that there is a HC excluding 1 from H1 (setti 4 conflict).

I m stuck at 22 solved cells (6 less than BP) and am still pretty sure, there is a chainless solution. The question is, who´ll find it.
@TRP: Thanks for your hint. Knowing it, leads me to your conclusion. But I still believe in an easier path...

@ Pets & BP

Look for a diagonal based SI in the bottom left.

@Leren: Do you mean the one that removes 1 from GHJ1? I've found that already. Or ist there another? Continue looking :-)

@pets: To be honest I used something that probably is a HC to get to 28 cells. At one point I had: If H8=1 then J1=1 and J3<>1, which leads to a setti conflict on 1.

@ Pets & BP

It's essentially the same, although not Setti based.

2 r8c2 => 4 r9c2 => 2 r9c1 => - 2 r8c2

I used that at 23 solved cells and that advanced me to 47 solved cells with singles.

Might be time to list your solved cells and I'll see if I can help.

@Leren: Okay, I started from scratch. Completely without HCs I got to 22 solved cells, like pets. Can you tell me how you found your 23th cell? Without it your SI does not work for me.

My solved cells are:

A19
B125689
C89
D45
E56
F238
G358
J78

OK we are cross posting here but here goes anyway.

Below, I've detailed every diagonal based move in my solver's solution. Somewhere in there you may find an elimination you've missed or that is new for you. This will also act as a tutorial for others. I've checked it twice, but there still may be typos. Of course there are plenty of other moves, but none is X based.

Solved Cells = 0
/ Pair of 9's A1, H8 => - 9 A2, H9
\ Symmetrical triple of 1's B2, E5, H8 => - 1 B8, H2
/ Pair of 1's E5, J1 => - 1 E1
/ Single 9 in A9
Solved Cells = 1
\ Single 9 in B2
Solved Cells = 8
\ Pair of 7's in AC1 => - 7 C3
Solved Cells = 9
\ Single 7 in A1
Solved Cells = 13
\ Naked pair (56) C3, J9 => - 5, E5, F6, G7, H8, - 6 D4
\ Pair of 2's DE5 => - 2 D4
/ Pair of 8's EG3 => - 8 E5
/ Pair of 8'S F4, G3 => - 8 G4
\ Pairs of 6's CE3, J69 => - 6 E6
\ Pair of 8's D4, F6 => - 8 F4
Solved Cells = 14
/ Pair of 5's C7, H2 => - 5 C2
\/ Aligned pairs of 5's C3, J9 & C7, H2 => - 5 H9, J2
\ X Wing of 8's D4, F6 + E46 => - 8 A4, HJ6
Solved Cells = 15
\ Pairs of 5's CE3, J69 => - 5 E6
Solved Cells = 18
\ Pair of 2's F6, G7 => - 2 G6
/ Pair of 4's F4, H2 => - 4 F2
/ Pairs of 2's GJ1, H29 => - 2 G9
Solved Cells = 19
\ X Wing of 8's D4, F6 + DF2 => - 8 A2
Solved Cells = 23
\ X Wing of 3's F6, H8 + FH9 => - 3 E9, G79
/ Finned X Wing of 3's C7, J1f, + E47 => - 3 J4
/ SI 2 H2 => 4 J2 => 2 J1 => - 2 H2
Solved Cells = 47
/ Finned X Wing of 2's C7, J1f, + E47 => - 2 J4
Solved Cells = 49
/ Finned X Wing of 3's C7f, J1, + G16 => - 3 C6

@ BP

OK, while you are cogitating over all those X moves I looked at your last post and noted that you don't have F4 solved.

After a basic clue markoff, F4 has 23489.

9 is excluded by the 1 in F7.

8 is excluded at 13 solved cells by \ Pair of 8's D4, F6 => - 8 F4.

At 18 solved cells I have H1 = 14 and H2 = 245, CE => H2 <> 4, leaving F4 = 4 as a single on the / diagonal.

Did that help ?

Wow, Leren, thanks a lot for the detailed info. Will check it out later and get back to you!

@ BP

The one thing I missed in the big X move list was the / single 4 in F4 at 18 solved cells.

I agree, I learned a lot from Leren's detailed procedure...some new moves for my repertoire! Thanks!

What is called a finned X wing and also (IIRC) the aligned pairs above I believe fits the description of an X-cycle in Sudoku.

@ Jan

Yes, the finned X wings, aligned diagonal pairs, and row/column pairs both touching the same diagonal are all examples of 4 cell discontinuous X chains, which you can't get in normal Str8ts. In normal Str8ts you can only get continuous loops, the simplest being an X wing.

So, I worked through your notes, Leren. Very interesting! I found out that I missed two things: The alignment of 5's that gives F4=4 and 3 in H89 and the skewed xwing of 3's in F6H8 + FH9. (I've never seen one like that, rather clever!!!) However that still does not help me with the SI. Cannot see why H2=2 would imply J2=4. In my puzzle J2=1 would be possible also. Probably am missing something obvious ...?

Also, could you explain what you mean by 4 cell discontinuous X chain?

@ BP

Also at 23 cells there is an ordinary X wing in 4's in AJ23 that => - 4 DG4, so in Column 2 only Rows & J contain 4.

Also in Column 2 only AH2 contain 5.

Then H2 = 2 is not 5. A2 = 5 and not 4, so J2 = 4, J1 = 2 and H2 <> 2. So if you want to be pedantic, there are really 4 cells involved in the SI.

I'll explain the X chain thing in another post.

Puh, looks rather complicated! I didn't read everything, so perhaps someone already found the same way as me, which was:
If H2 wasn't 5, there'd be no 5s left in that row (via C7, C3, J1..), so H2 must be 5.

To familiarise yourself with continuous and discontinuous X chains go to Andrew's Sudoku X Cycles section and just read the parts that describe continuous and discontinuous X loops.

What Andrew doesn't mention (and it took me quite a while to realise, is that (in Sudoku) discontinuous X loops always contain at least one "diagonal" link in a box. In Str8ts X the diagonals act like the boxes do in Sudoku, in allowing discontinuous X loops to be formed.

Let's take the following example from this puzzle.

/ Finned X Wing of 3's C7, J1f, + E47 => - 3 J4

You could view this as a discontinuous X loop as follows.

Assume J4 = 3. Then E4 <> 3 (a Weak link). So E7 = 3 ( a Strong link), C7 <> 3 (a Weak link), J1 = 3 (a Strong link on a diagonal) and J4 <> 3 (a Weak link).

So the loop starts and ends on Weak links, which makes it discontinuous. As I've already said, you can't get discontinuous X loops in ordinary Str8ts because you have only rows and columns to work with. In ordinary Str8ts, All fish are ultimately continuous loops.

One more thing about discontinuous X loops. Some people describe them as "contradiction" chains but that's not strictly true.

So, in our example we assume J4 = 3 and conclude (via some argument), that J4 <> 3. But if you also assume J4 <> 3 then J4 <> 3. Now that may be obvious, but it's an important part of the proof that J4 <> 3. Why is that ?

Well, what we have shown is that if J4 = 3 and J4 <> 3 then J4 <> 3. Since J4 can only be 3 or not 3 we have covered all our bases, and are entitled to conclude that J4 <> 3 (the common outcome).

That's a constructive proof, not a contradiction argument.

I don't mind people calling these things contradiction chains but it's not strictly true (and I always have a bit of a chuckle to myself).

@Leren: Okay, now I get it! Thank you for your patience! All this to avoid a tiny HC ... :-)

And thank you for your explanation about X chains. I'll tackle that later, now my brain needs a good long break!!!

I knew there's a way through.
Thank you, Leren for the brilliant explanation - I hope I find time to get your strategies!

is.gd/Leren_500

thanks Leren, standard tools and a head chain for the end

Same for me. Thank you, nice puzzle.

Very enjoyable. Standard tools and UR to avoid the end chain. Thank you!

Very interesting; good degree of complexity.

If you choose H6=4, you get a "solution" that only has 8 instead of 7 in row, all the rest is consistent.

WIth 21 solved cells after the 4-wing, I can make use of the UR that solves J3; after some further wings, I need either the second UR based on the many (12) cells or a short chain disproving H6=4.

Is that what you did as well...?

Reserved for discussion of Andrew's puzzle.

I found this very chainy without extra clues. Guessing the right value in F5 or F6 makes it too easy. The best I came up with was add A5 = 5 and G2 = 2 and you get a reasonable Extreme.

I couldn't solve this puzzle on my own, so I read Leren's comment. I was surprised, that the compartment F56 provides an easy solution. It has three different str8ts as candidates: 12, 56 and 89 with 56 and 89 bound to compartment F12. In addition each of the candidates 5, 6, 8, 9 is present in only one of the two cells. The question is which of the three str8ts is the solution in F56. I choise the option that made it easier to construct the path. There is some logic in the situation's assessment and choice of action. It seems to me it isn't only guessing.