Try these Orange Bars!

The grid has follwing properties: 1. Disjoint groups rules apply. 2. Every row and column forms valid "roping" pattern. 3. The lines are all also renban lines and modular lines. 4. If you swap the rows 4 and 5 (i.e. for each column, swap the digits in row 4 and row 5), the grid will be 180 degrees rotationally symmetrical and still obeys disjoint rules.

Tomato Pie here, and wow this was delightful to wake up to. Thank you so much for featuring my puzzle! It's genuinely a dream come true. Great solve Mark I especially loved watching you discover how the clues in box 9 disambiguate the puzzle as that was exactly how I intended it. The reason I used yellow lines in my original puzzle is because I typically recognise orange lines as modular. The purpose of the >1 skyscrapers was to force the 3,6,9 triple into the middle column of rows 1,2 and 3 so I apologise if that could have been clearer. Thank you so so much for featuring my first ever puzzle. P.S. congrats on your bonus points🎉🎉, modular and renban is correct.

Mark's proof that all rows are cyclic because no non-cyclic rows exist is not correct. Valid non-cyclic rows: 591627384 592617384 593716284 593726184 594816273 594826173 594837162 615927384 615937284 615948273 481593726 371594826 627159384 627159483 I believe that these, together with their 10s complements (i.e., switch 1s for 9s, 2s for 8s, etc.) and the reverses of each are the exhaustive list of valid non-cyclic sequences. All that said, it is definitely not necessary to prove cyclicity to solve this puzzle.

Interestingly I mistook the sum in the middle 3 rows for a thermometer- Which also had a complete unique solution!

I'm not embarrassed to say I was only able to finish it using a spreadsheet.

Wow. Those clues look really, really weird. Edit: Cycling in Dutch Whispers is highly appropriate for a biking country.

Season 3 is soooo good. The Gus scenes in this episode are great and Mike fixing Chuck's door is one of my favourite scenes in the whole show. I'm very excited for the next reaction knowing what episode is next.

Unrelated, do people actually make orange bars? Because I love lemon bars and imagine orange would be delightful.

Did it in just over 9 minutes once I realized what the lines were , by using "place holder logic" that makes things clear. The top just fills in, with a possible need to flip it if the order is wrong. The arrow then lines up the middle 3 rows (using the non-obvious other line property), and then the inequalities the bottom 3 rows. The one weakness of the method is that since I guessed right at the beginning I didn't verify that the solution was unique, I would have had to reverse the top and find where things don't work. As others have said, when you think of it as a miracle puzzle, it sort of solves itself.

I'm about an hour into trying to solve and just have a giant list of about 14 possible rows (28 with reversal, 56 with ten's complements), and many are not cyclical, which Mark seems to discount entirely. Other than the 2 skyscraper row having only two mirror options, I'm at a loss for how to continue. Mark handwaves away the logic too much here. I'm not a fan of this one as it seems like you're meant to just throw numbers at it until it works instead of logic it out. I haven't yet arrived at logic that doesn't need cyclical proof to solve and I'm too exhausted to prove it correctly. This puzzle has defeated me thoroughly.

61:40 for me. The way I "proved" to myself that each row needed to use the cyclic order of digits was to consider the limited-partner digits (4, 5, and 6) and what would happen if you didn't stick them "where they're supposed to go" (can't think of better wording right now). Imagine you have a 159 somewhere in the middle of a row. The 4 is supposed to go beside the 9 between it and an 8. If you instead decide to not put the 4 there, the only other location it could go would be on the leftmost or rightmost cell of that row; suppose you place the 4 on the rightmost cell. Now consider another 159 in another row with the 9 in column 8; the 4 should be placed in column 8 per the "cycle", but there's already a 4 in that column, so we have to stick it on the other end (which wouldn't actually work because of polarity, but ignore that for now). Now, suppose in a third row you have 159 with the 9 in column 9. Where can you place the 4? You can't put the 4 in any cells with two neighbors because you can't use the 9, and it can't go in column 1 because there's already another 4 in that column. The takeaway is that if you separate the 4 from the 9 and break the cycle you will necessarily break another row later on. The way to avoid this is to not split up the 4 from the 9 and keep all the digits in a cycle. You can repeat this logic for the other two restrictive digits (5 and 6). There's probably a better way to prove this, but I can't think of anything besides case testing and looking forward three or four steps.

The title might be a vague reference to "double dutch", which refers to a game of jump rope where two people stand on either side of the player turning a pair of lengthwise jump ropes around him or her.

Has anyone found a more elegant and complete way to prove the correct sequence of digits in this puzzle? I consider proving the logical moves to be part of the solve, so this solve wasn't entirely satisfying.

One of the best puzzles I have managed to solve. Very satisfying logic.

Nice miracle sudoku. I didn't robustly prove the lines had to be cyclic. Just assumed they were. Once I determined the cycle it was just a matter of marking things out appropriately. Finished at what felt like a snail's pace of 33:00. For those wondering, the lines were also renban and modular.

Its also fascinating that the digits kind of corkscrew in ascending order through the three sets of three rows as well

A Proof of Cyclicity 1. Column 3 must contain a digit 5 somewhere. Let us call the row in which five appears in the 3rd cell, Line 3 (L3). 2. The digit five must be flanked by one and nine. As the logic that follows is symmetrical around pairs of digits summing to ten, we can consider the case 1-5-9 without loss of generality. 3. The cells neighbouring 5 in L1, L5, L7 and L9 are all now determined. 4. We can place 6 in all these lines, starting with L3, by considering a) polarity, b) six’s possible neighbours, and c) vertical sudoku. 5. Similarly, 4 is placeable starting with L7. Then 8s, as they are the only other legal neighbours for 4; then 3s, etc. All of the odd lines L1,3,5,7,9 are fully disambiguated. 6. By considering vertical sudoku, 5’s left-hand neighbour in L2 is disambiguated; which cascades to the rest of the grid. 7. The pattern is seen to be cyclic - QED.

31:25 A mathematicians dream is my nightmare lol

14:11 Nice and simple, but cute and elegant. The whisper lines double (and triple) as region sum lines, and modular lines.

I don't know that it changes anything but there are 72 unique rows that work, 56 if you don't count the cycles. One ignored case can be seen at 13:22 where it starts with 2,7,3,8 which could also start with 3,7,2,8.