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.
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.
But, for example, taking the 5 on the end to not need a 1 and 9 on the cyclic positions puts the 1 for example in a column that the in other two other rows will have a 5 in an adjacent column, needing either a 1 or a 9 in this column giving you 3 1/9s in just one column. Columns 2 through 8 all need their ones and nines to handle fives, so no spare space for the one not needed by the five on the end,
@@RichardDamon I'm not saying the solution Mark found is wrong (or that it's not the only solution). I'm saying he sidestepped the logic needed to prove that every row uses the cyclic sequence by falsely claiming that there are no valid non-cyclic sequences for a single row. (Also, what you're arguing only rules out half of the non-cyclic sequences directly, though there may be an extension using the 4s and 6s in addition to the 5s that gets the full way there.)
@@RosefinchCarr As Richard already pointed out, the fact that eight 1s and eight 9s (including all the ones in columns 2-8) have to be next to 5s does quite a bit of work. And, the inequalities in box 9 do more than they seem like they should. Probably the biggest thing is that, by counting high digits, they force r7c8 to be high and not be next to a 5. From the point above, that ends up putting a 9 into one of r89c7 with no 5 to its right. Ultimately, doing some additional work with row 7, given what can be determined about 159 in boxes 8 and 9, together with the restrictions from the skyscrapers lets you place every 159 into a vertical triple at worst. Then, the whisper in r7 and the combination of whispers and skyscrapers in r123 is enough to finish off everything but the middle three rows, which are done by the arrow.
@@SSGranor I don't agree with you that Mark would have claimed there is no single valid non-cyclic sequence. I think he claimed that you cannot fill the grid with sequences like that because of what Richard commented above. I do agree he doesn't prove that.
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.
Spoilers ahead: also my way of proving the repeating sequence was to just start placing 5s in each row and column, and then 1s and 9s. Because digits see each other in the columns you can quickly see that only one sequence works (or the reverse of it). So for anyone looking for a more complete/satisfying proof, the secret is in the columns.
@@RosefinchCarrThank you for the explanation of your method of proving the cyclic requirement. I admit, I initially solved it by assuming the toroidal cyclic pattern was probably what was required, but I wasn't very satisfied with my approach. I then watched Mark's solve, and was still left unsatisfied. But following your comment, I tried putting 5's down the negative diagonal, deciding I'd accept the rows would need re-ordering later, but let's just see where applying purely column logic gets us, ignoring row position for now. That allowed me to prove the rows were indeed the same pattern, just cycled. Cheers.
@@RosefinchCarr Great puzzle! I noticed that once you figure out the repeating sequence in the way you described, you can even do without both the >1 skyscraper clues.
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.
Thanks very much! That’s a nice proof, just try solving the puzzle- put 5 in the center, then in the third and seventh columns, and just keep shifting the 5 around the rows! Nifty!
@@laurv8370 Reversing the order in some rows doesn't work because 5's will force the neighbors of 5's two columns away (that is, all the even column 5's have to use the same order as each other, and all the odd column 5's have to use the same order as each other), and reversing the orders between the two sets of columns causes a clash between one of the 5's and its neighbor. First, to show that the 5's force the neighbors of the other 5's two columns away, take the 5 in Column 2. It will need the 1 and 9 around it in Columns 1 and 3. For this example suppose the 9 is to the right in Column 3 (so 159). The 5 in Column 4 then can't have a 9 in Column 3, it will need to be a 1 to the left and the 9 to the right. This follows for all 5's in even columns so they will all be 159. Now consider the 5's in the odd columns (skip Columns 1 and 9 for now since those 5's have only one neighbor each) and suppose we reversed the order. The 5 in Column 3 will again need a 1 and 9 on either side of it; this time we put the 9 to the left and 1 to the right in Column 4 (so 951). Knowing that this order is forced for the 5's in the other odd columns (per the previous paragraph), we then know that the 5 in Column 7 has a 1 to the right of it in Column 8. Thus the 5 in Column 9 can't have a 1 to the left of it in Column 8, so it will need a 9 to be there. Now there is exactly once cell left in Column 1 where both 5 and 9 will have to fit, causing a clash. In any of the rows where 5 has two neighbors (Columns 2-8) there's a 9 already in the row. In the row where 5 is in Column 9 there's again a 9 already in the row. The only row in Column 1 that doesn't have a 9 in it is also the only row that doesn't have a 5 in it, thus placing either digit in that cell will prevent the other digit from being placed in that row or column. Here's a crude mockup of what the rows would look like, lined up as best as I could make a RUclips comment look. The X in the bottom left is the only spot for both 5 and 9 in that row and column. If you put a 5 there you couldn't place the last 9 anywhere in the row or column (likewise for placing 9 there, you couldn't then place the last 5 anywhere). To fix this you would need to use the same order for every row, thus having a mix of 159 and 951 doesn't work. 1 5 9 | _ _ _ | _ _ _ _ _ 1 | 5 9 _ | _ _ _ _ _ _ | _ 1 5 | 9 _ _ _ _ _ | _ _ _ | 1 5 9 _ _ _ | _ _ _ | _ 9 5 _ _ _ | _ _ 9 | 5 1 _ _ _ _ | 9 5 1 | _ _ _ _ 9 5 | 1 _ _ | _ _ _ X 1 _ | _ _ _ | _ _ _
@@RecreationalCynic for me it is very clear, and also clear why it is cyclical and it only goes one direction, but it was not explained like that, that's why I commented
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.
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.
The following works for the numbers 4,5,6, and once theyre in place you can fix the rest. I'll use 5 as an example. Imagine 7 of the rows that have 5 in columns 2 to 8. Now consider the 5 in column 1, the cell in column 2 next to it must be 1 or 9, and we can see it must be the one that is always on the right, because there is a somewhere a 5 in column 3 and it has a 1or 9 in col 2 next to. So back to the row with 5 in column 1, the only place the other 1 or 9 can go is in column 9 - thus showing that the 159 (at least) are in the same order for every row. Add in the 4 looking at 8 or 9 and the 6 looking at 1 or 2 and we get that the 2615948 must cyclic round for each row. Then the 3 and 7 only have 1 option but to fit in.
Yes, if you consider this more elegant. Logical explanation: Sequence 2-6-1-5-9-4-8 is not flexible, as you cannot replace any of the digits around 4,5 and 6 or even switch them around as all those 3 digits can only go between two specific digits. All you could do is cut that sequence by using the edge of the grid. However there are 2 rows where the leftmost edge digit will be a 3 or a 7 (using left just for example), which will cause the sequence to exist in its entirety. Then the OTHER row that starts with 3 or 7 cannot just reverse that sequence since the middle digit of both of those rows would be the same, which means the next digit on that row has to be the other one from 3 and 7. That also A) forces the sequence to move in the same direction and B) it forces the row where 3 and 7 are in the ends of the row to have a 159 sequence (either direction) in the middle box. That will dictate the position of 4 and 6 to be on the edges of that box, as their other counterparts (1 and 9) are already in the box. Again, if you would try to reverse the order, the sudoku will fail you very fast so the edge where 4 and 6 must be is actually determined. Once you hypothicate which row has the 4 and which has the 6, you can actually pencilmark it to prove that it forces roping. That was a long and winding way to prove not only that the sequence exists, but it also moves the same way and causes roping in every row. (as Proof of concept method: Place all 5s (or 4s or 6s) to the grid in any arbitrary positions as long as you obey sudoku rules. After that surround them completely with 19 middle pencilmarks. As soon as you fill a single digit anywhere it will immediately populate all 1s and 9s including the ones that go over to the other side (where 5 is on the edge). Then those 2 rows where 5 is on the edge will force the position of 6 and 4, 2 and 8 and so forth, until the whole grid is populated simply by enforcing sudoku and dutch whisper rules. If you go back to the point where you decided between a 1 and a 9, you can reverse the decision by using the other digit, and the grid will again fully populate but just in reverse order.
@@MarcusTheDorkus No it isn't, you can solve the puzzle without assuming anything using only the given rules and sudoku. In fact trying to "prove" it using trial and error, which is what most people who "proved it" did is actual bifurcating.
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.
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.
I encourage you to keep trying, Trust me there is a logical way to prove that there is only one possible sequence that works and the only rules you need to prove that are dutch whispers and normal sudoku rules. If you need a hint then I would say that all 9 rows are needed to prove it (the secret is in the columns).
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.
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.
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.
I solved from the bottom right corner, deducing the direction of the sequence from the grey square and < interactions. From there, the middle 3 rows can be deduced in the same way. I ended up with the bottom 6 rows done and then had no idea what the skyscraper rules meant and had to watch the video to understand. The rules are not well written.
I didn't think of the modular lines. But for the second option: If you mix a little red colour into your blue region-sum-lines, then they'll become what they really are.
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.
At 14:36, Mark missed that there is a a set of valid row beginning with 62 and ending with 15: 6 2 [789] [34] [89] [34] [789] 1 5. These rows do not wrap, of course, but you must prove why they do not work anywhere.
14:31. I first figured that row 3 had to have 4 and 8 at the ends to get a double 2 sky scraper, giving me 6 in the middle. Row two needed to have 2 and 7 at the ends to get double 3, giving me 9 in the middle. Then I figured the only way to make the arrow work was to make it 1+2=3, which gave me the direction. Final I figured the order of the bottom three rows with the and that the square had to be even and > 2
Finished in 45:41. Interesting way the Dutch lines interact with each other and particularly the skyscraper clues. It was easy to make a bad assumption, but since the places are so limited, it makes it rather simple to backtrace the errors. Fun puzzle!
Worked it out entirely in a notepad app beforehand. Then just filled it in. Wrote out the only sequence that works, just had to figure out which direction it went and where it started on each row. For the top 3 rows, I figured out where each sequence started in each row, but not the direction. That told me that each row among those 3 must shift by 1 for another set of boxes and another 1 for the final set of boxes. There was only 1 option within those that worked for boxes 7, 8, and 9, which told me the direction (for every row) and also which set was in each row in the bottom. That gave me rows 1, 2, 3, 7, 8, and 9. For the middle row, it was pretty simple to find the only possibility for the order of the rows using the arrow clue. Whole thing took 14:51.
I usually don't read comments much before leaving one myself, but this time, because you invited folks to give other logical approaches, I was curious to see what folks said. Maybe there are others? I can see that there may be other sequences. But I don't think that anyone has proven either your sequence to be the right one nor that there are not more than one - but I am not really reading carefully at this point, just skimming to satisfy my curiosity in a general way. Nonetheless, it was fascinating to hear you reason out how it all might work, and how you came to the correct solution. I will not be trying this puzzle myself, but I am very glad to have had the chance to watch you do it. By the way, "torus" and "toroidal" are words I had never heard before watching this channel. An education on so many levels. Thanks, Mark, as always.
I loved that. Had to think a bit to get started in box 9. Then I decided that the 2 rules rule would have to be significant. Renban was obviously 1 rule and then some sort of cycling. Quickly that was obviously modular. I was slow to notice that each column in a box was strictly 1 modulus. Thereafter it fell out very quickly.
Not really a 'solve' if you just managed to find a way (lucky or not) that filled in the grid in a valid way. Why? Because you have not demonstrated that your biffed solution is unique. Still - if you are happy to not work through the logic - and assume uniqueness - then who am I to challenge? ;-)
@@schubertuk Actually I'm taking it the opposite way around: If a puzzles solution is not unique, then, depending on the way you see it, either it is not solvable (thus an unfair win to an unfair task is fair game) or any solution counts. Vice versa, IF the puzzle must solve uniquely, that is a rule in itself that can be utilized.
An easy proof of why the the rows of dutch line must be cyclic (Open the puzzle for easier understanding) (we say 123456789 and 912345678 and 891234567 are cyclic) Things to know: 1. 5 can only be next to 19, 4 next to 89, 6 next to 12 on the dutch line 2. After we put in 5, everything else alternates between big (6789) and small (1234) Preparation: Put all 5's in the diagonal, r1c1,r2c2,r3c3... We will ignore the boxes in this proof. Just look at the rows and columns Step 1: What will happen when 5 is in column 2? Assume it is 159xxxxxx, as everything alternates. It will be 159sBsBsB, notice that 6 must be next to 1 or 2, so it must be at the end as 26. 4 must be next to 89 so it will be 948. Result: 159483726 is the only sequence. (If we start with 951, it will be 951627384, we will now refrain from talking about this second possibility, as it is just swapping 1-9,2-8,3-7,4-6) Step 2: What will happen when 5 is in column 4,6,8? Notice that row2 is now 159483726, column 3 cannot have another 9, so row4 must be xx159xxxx, continue with this logic. row 2,4,6,8 will all have 159, not 951 Step 3: We can now fill in everything in row 2468, and it will be cyclic. Hint: row 8 can be filled in a way similar to step2. (159483726, 261594837, 372615948, 483726159) Step 4: 5 in column 3,5,7 Notice now column 1 already have 1234 (small), so if column 2 is big, column 1 has nothing to be placed. It cannot be x951xxxxx, it must be x159xxxxx. Repeat the process in Step 2, row 3,5,7 will all have 159. Now fill in all the 1's and 9's (row 1 will be 59xxxxxx1, row 9 will be 9xxxxxx15) Step 5: Fill in everything else. (hint: fill in row 19 first, then look at column 4,6) Remember the two things on a dutch line! Conclusion, All rows on the grid must be cyclic
Has anyone proven that Mark's is the one and only solution? Seems like a huge shortcut was taken in dismissing non cyclical rows which happens to work because that's what the author intended.
I'm irrationally annoyed by how they've got the usual "These Two Cells Add To Five" V symbol there instead of the properly wide inequality symbol to mean "R8C9 is bigger than R9C9".
I’m seeing a bit of discussion about how to prove the cyclic sequence so I decided to try it on my own and this is what I came up with (I did see one comment by RecreationalCynic which spurred my thought process so the credit for determining that the cycles all go in the same direction is theirs and not mine) First, think about the 5 in all of the even columns and assume that the cycle goes 159 for the 5 in c2, the 9 being placed in c3 means that the 5 in c4 must have a 1 to its left in c3 which means that the order of the sequence in one even column will force the same order in at least one other even column 5 which will then propagate and force the order 159 in all rows where the five is in an even column (this was the point where I stopped reading their comment and decided to approach it for myself, so I don’t actually know how they proved the order must be the same for all the 5s in an odd column) Next, consider the 9 in c1, because we have already placed a 5 in c2 there can’t be another 5 in that column meaning 5 only has one neighbor (1), and must go in c9 with a 1 in c8, this then forces the order in a row where the 5 is in an odd column so the effect again propagates and now the same 159 order gets forced in all rows where the five is an odd column, and the same order being forced wherever there’s a 5 in any even or odd column just means that it’s forced in every column The next digit to consider is the 6 in the row where the 1 is in c1, because it can’t go next to the 1 it only has one neighbor (2) placing it in c9 and being that it’s the only 6 that can be there every other 6 goes next to a 1, this can then place all of the 2s next to them The same logic can be used to place a 4 in c1 where the 9 is in c9, which can then place all of the 4s and then 8s Then the 3s can be placed next to the 8s and the 7s next to the 2s This completely ignores row order, but it does prove that every row must be that same sequence just shifted
So close to pairwise solution but the cluster of rules in B9 obviously breaks the duality... Did not think pairwise could exist w arrow sum but there existed a 2nd sum. If the C9R89 < did not wxist could probably form a pairwise.
That arrow was easier to solve if you use already written 3 rows and search there possible location for that arrow and then you see that diagonally these digits are consecutive so that reveals these must be 1 2 3 there as using 7+8 = 9 is nonsensense. Or 8+9=1
Once you understand the way the grid needs to be laid out, actually filling in the digits is easy. I got an 8:26 on this puzzle, but that doesn’t include the 10+ minutes I spent figuring that out.
49:29, spent like 5 minutes not seeing how to go, then immediately had an idea on my walk and started pencil marking when I got back home. My solve was basically going "if this was a 9" in rows 1, 2, and 3, pencil marking everything, then reversing all that work in case it was the other direction. Managed to solve from there, but it was little logic, lot of trial and error.
I get the bonus points -- renban is obvious, and the other is modular (mod 3 rules). Vertically, each column was entropic in that each block had one from low, medium, and high. Also vertically, each block had the same modularity. Roping was everywhere, both vertically and horizontally. Does this make up for me taking forever to do it?
_"It's sort of a prove by example"_ (Mark @13:50) Unless you provide a method to generalize, that prove is only valid for that particular example. I used the other method, as explained by *Tomato Pie.* I doubt your example can be generalized, as it does not consider interactions with other rows. I doubt this is enough to prove *cyclicity.*
I enjoyed solving the first three rows of this puzzle, but after that it got a bit frustrating. I could see that it was going to be the same cyclic sequence in every row - probably with roping - but couldn't prove it. I ended up writing out the full contents of the remaining six rows and then rearranging them until the other clues were satisfied. I would love to know if anyone has found a more elegant route to the solution.
Things to know: 1. 5 can only be next to 19, 4 next to 89, 6 next to 12 on the dutch line 2. After we put in 5, everything else alternates between big (6789) and small (1234) Preparation: Put all 5's in the diagonal, r1c1,r2c2,r3c3... We will ignore the boxes in this proof. Just look at the rows and columns Step 1: What will happen when 5 is in column 2? Assume it is 159xxxxxx, as everything alternates. It will be 159sBsBsB, notice that 6 must be next to 1 or 2, so it must be at the end as 26. 4 must be next to 89 so it will be 948. Result: 159483726 is the only sequence. (If we start with 951, it will be 951627384, we will now refrain from talking about this second possibility, as it is just swapping 1-9,2-8,3-7,4-6) Step 2: What will happen when 5 is in column 4,6,8? Notice that row2 is now 159483726, column 3 cannot have another 9, so row4 must be xx159xxxx, continue with this logic. row 2,4,6,8 will all have 159, not 951 Step 3: We can now fill in everything in row 2468, and it will be cyclic. Hint: row 8 can be filled in a way similar to step2. (159483726, 261594837, 372615948, 483726159) Step 4: 5 in column 3,5,7 Notice now column 1 already have 1234 (small), so if column 2 is big, column 1 has nothing to be placed. It cannot be x951xxxxx, it must be x159xxxxx. Repeat the process in Step 2, row 3,5,7 will all have 159. Now fill in all the 1's and 9's (row 1 will be 59xxxxxx1, row 9 will be 9xxxxxx15) Step 5: Fill in everything else. (hint: fill in row 19 first, then look at column 4,6) Remember the two things on a dutch line! All rows on the grid must be cyclic
I think that it has been used to fix which rotation of the sequence must be in each row, based on how many digits can be "seen" from each side of the puzzle. Without them I am not sure how one would determine that. At least, that is how Mark used them.
Without the skyscraper clues, you could take Mark's solution and swap rows 1,2,3 and obtain alternative solutions. The skyscraper clues force uniqueness.
Rules: 02:31 Let's Get Cracking: 04:59 And how about this video's Simarkisms?! Sorry: 7x (09:31, 09:44, 10:19, 12:44, 17:10, 22:09, 29:32) Obviously: 3x (11:13, 29:23, 33:07) Bother: 2x (20:46, 23:29) Fascinating: 2x (01:01, 04:41) Goodness: 1x (30:53) Nonsense: 1x (32:28) Clever: 1x (27:46) Approachable: 1x (01:54) Hang On: 1x (25:29) Surely: 1x (32:31) Proof: 1x (14:32) In Fact: 1x (13:24) Intriguing: 1x (26:53) Weird: 1x (27:41) Most popular number(>9), digit and colour this video: Ninety Four (7 mentions) One (57 mentions) Orange (6 mentions) Antithesis Battles: High (2) - Low (1) Even (9) - Odd (0) Outside (2) - Inside (0) Row (41) - Column (7) FAQ: Q1: You missed something! A1: That could very well be the case! Human speech can be hard to understand for computers like me! Point out the ones that I missed and maybe I'll learn! Q2: Can you do this for another channel? A2: I've been thinking about that and wrote some code to make that possible. Let me know which channel you think would be a good fit!
On that exact sequence, yes you can, but using the sequence as a torus where is wraps around to the beginning, the 3 and 6 would be next to each other. That's a difference of 3, not 4.
@@pacman52280 The grid isn't a torus. Mark is trying to prove that, nonetheless, the toroidal sequence must be used in every row; and, he ends up doing so by claiming, falsely, that there are no other valid sequences for any row.
In his row 4, that has 5 in the middle (726159483), you can also swap 7 with 6 and/or 4 with 3 too. I ended up assuming that it would be the toroidal pattern cycling and just seeing if that worked. I wasn't surprised that it did, but I knew I hadn't proved uniqueness. I was looking forward to seeing how Mark would achieve his step (B) once he mentioned it as a necessary step in the proof. Most disappointing that he didn't achieve this.
Did not like this puzzle at all. I sort of tried to play around and see "what happens if we put this here" and how it affects other things. Started in the middle and the arrow accidentally lined up and worked, so the middle 3 rows were correct, then tried to see what would satisfy the skyscraper clues and then the bottom clue and whoops - it all solved and the solution "was correct", but I had no idea why it's unique and how to prove it. Watched the video and turns out Mark just remembered if from somewhere else and sort of "saw" that it would work here. Not very satisfying - I would have liked a full apriori logic of how to deduce this from scratch and why it was necessarily the only solution.
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.
Good catch on the disjoint grouping !
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.
But, for example, taking the 5 on the end to not need a 1 and 9 on the cyclic positions puts the 1 for example in a column that the in other two other rows will have a 5 in an adjacent column, needing either a 1 or a 9 in this column giving you 3 1/9s in just one column. Columns 2 through 8 all need their ones and nines to handle fives, so no spare space for the one not needed by the five on the end,
@@RichardDamon I'm not saying the solution Mark found is wrong (or that it's not the only solution). I'm saying he sidestepped the logic needed to prove that every row uses the cyclic sequence by falsely claiming that there are no valid non-cyclic sequences for a single row.
(Also, what you're arguing only rules out half of the non-cyclic sequences directly, though there may be an extension using the 4s and 6s in addition to the 5s that gets the full way there.)
Tomato Pie here, as the setter of the puzzle I didn't know it was possible without proving the repeating sequence, could you tell me more?
@@RosefinchCarr As Richard already pointed out, the fact that eight 1s and eight 9s (including all the ones in columns 2-8) have to be next to 5s does quite a bit of work. And, the inequalities in box 9 do more than they seem like they should. Probably the biggest thing is that, by counting high digits, they force r7c8 to be high and not be next to a 5. From the point above, that ends up putting a 9 into one of r89c7 with no 5 to its right. Ultimately, doing some additional work with row 7, given what can be determined about 159 in boxes 8 and 9, together with the restrictions from the skyscrapers lets you place every 159 into a vertical triple at worst. Then, the whisper in r7 and the combination of whispers and skyscrapers in r123 is enough to finish off everything but the middle three rows, which are done by the arrow.
@@SSGranor I don't agree with you that Mark would have claimed there is no single valid non-cyclic sequence. I think he claimed that you cannot fill the grid with sequences like that because of what Richard commented above. I do agree he doesn't prove that.
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.
Spoilers ahead: also my way of proving the repeating sequence was to just start placing 5s in each row and column, and then 1s and 9s. Because digits see each other in the columns you can quickly see that only one sequence works (or the reverse of it). So for anyone looking for a more complete/satisfying proof, the secret is in the columns.
Somebody pin Tomato Pie's comment 😆
Good job on the puzzle set 🎉 Hoping to see more of your puzzles featured in this channel ❤
@@zariahzariah1391 Thank you!
@@RosefinchCarrThank you for the explanation of your method of proving the cyclic requirement. I admit, I initially solved it by assuming the toroidal cyclic pattern was probably what was required, but I wasn't very satisfied with my approach. I then watched Mark's solve, and was still left unsatisfied. But following your comment, I tried putting 5's down the negative diagonal, deciding I'd accept the rows would need re-ordering later, but let's just see where applying purely column logic gets us, ignoring row position for now. That allowed me to prove the rows were indeed the same pattern, just cycled. Cheers.
@@RosefinchCarr Great puzzle! I noticed that once you figure out the repeating sequence in the way you described, you can even do without both the >1 skyscraper clues.
Interestingly I mistook the sum in the middle 3 rows for a thermometer- Which also had a complete unique solution!
Wow. Those clues look really, really weird. Edit: Cycling in Dutch Whispers is highly appropriate for a biking country.
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.
Thanks very much! That’s a nice proof, just try solving the puzzle- put 5 in the center, then in the third and seventh columns, and just keep shifting the 5 around the rows! Nifty!
This assumes you have 159 in all rows, but how about some rows having 159 and some 951?
@@laurv8370 Reversing the order in some rows doesn't work because 5's will force the neighbors of 5's two columns away (that is, all the even column 5's have to use the same order as each other, and all the odd column 5's have to use the same order as each other), and reversing the orders between the two sets of columns causes a clash between one of the 5's and its neighbor.
First, to show that the 5's force the neighbors of the other 5's two columns away, take the 5 in Column 2. It will need the 1 and 9 around it in Columns 1 and 3. For this example suppose the 9 is to the right in Column 3 (so 159). The 5 in Column 4 then can't have a 9 in Column 3, it will need to be a 1 to the left and the 9 to the right. This follows for all 5's in even columns so they will all be 159.
Now consider the 5's in the odd columns (skip Columns 1 and 9 for now since those 5's have only one neighbor each) and suppose we reversed the order. The 5 in Column 3 will again need a 1 and 9 on either side of it; this time we put the 9 to the left and 1 to the right in Column 4 (so 951). Knowing that this order is forced for the 5's in the other odd columns (per the previous paragraph), we then know that the 5 in Column 7 has a 1 to the right of it in Column 8. Thus the 5 in Column 9 can't have a 1 to the left of it in Column 8, so it will need a 9 to be there.
Now there is exactly once cell left in Column 1 where both 5 and 9 will have to fit, causing a clash. In any of the rows where 5 has two neighbors (Columns 2-8) there's a 9 already in the row. In the row where 5 is in Column 9 there's again a 9 already in the row. The only row in Column 1 that doesn't have a 9 in it is also the only row that doesn't have a 5 in it, thus placing either digit in that cell will prevent the other digit from being placed in that row or column.
Here's a crude mockup of what the rows would look like, lined up as best as I could make a RUclips comment look. The X in the bottom left is the only spot for both 5 and 9 in that row and column. If you put a 5 there you couldn't place the last 9 anywhere in the row or column (likewise for placing 9 there, you couldn't then place the last 5 anywhere). To fix this you would need to use the same order for every row, thus having a mix of 159 and 951 doesn't work.
1 5 9 | _ _ _ | _ _ _
_ _ 1 | 5 9 _ | _ _ _
_ _ _ | _ 1 5 | 9 _ _
_ _ _ | _ _ _ | 1 5 9
_ _ _ | _ _ _ | _ 9 5
_ _ _ | _ _ 9 | 5 1 _
_ _ _ | 9 5 1 | _ _ _
_ 9 5 | 1 _ _ | _ _ _
X 1 _ | _ _ _ | _ _ _
@@RecreationalCynic for me it is very clear, and also clear why it is cyclical and it only goes one direction, but it was not explained like that, that's why I commented
Unrelated, do people actually make orange bars? Because I love lemon bars and imagine orange would be delightful.
I haven't had orange bars but I'm sure they exist. I have had grapefruit bars which were an interesting variation.
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.
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.
The following works for the numbers 4,5,6, and once theyre in place you can fix the rest.
I'll use 5 as an example. Imagine 7 of the rows that have 5 in columns 2 to 8.
Now consider the 5 in column 1, the cell in column 2 next to it must be 1 or 9, and we can see it must be the one that is always on the right, because there is a somewhere a 5 in column 3 and it has a 1or 9 in col 2 next to.
So back to the row with 5 in column 1, the only place the other 1 or 9 can go is in column 9 - thus showing that the 159 (at least) are in the same order for every row.
Add in the 4 looking at 8 or 9 and the 6 looking at 1 or 2 and we get that the 2615948 must cyclic round for each row. Then the 3 and 7 only have 1 option but to fit in.
Yes, if you consider this more elegant.
Logical explanation:
Sequence 2-6-1-5-9-4-8 is not flexible, as you cannot replace any of the digits around 4,5 and 6 or even switch them around as all those 3 digits can only go between two specific digits. All you could do is cut that sequence by using the edge of the grid. However there are 2 rows where the leftmost edge digit will be a 3 or a 7 (using left just for example), which will cause the sequence to exist in its entirety. Then the OTHER row that starts with 3 or 7 cannot just reverse that sequence since the middle digit of both of those rows would be the same, which means the next digit on that row has to be the other one from 3 and 7.
That also A) forces the sequence to move in the same direction and B) it forces the row where 3 and 7 are in the ends of the row to have a 159 sequence (either direction) in the middle box. That will dictate the position of 4 and 6 to be on the edges of that box, as their other counterparts (1 and 9) are already in the box. Again, if you would try to reverse the order, the sudoku will fail you very fast so the edge where 4 and 6 must be is actually determined. Once you hypothicate which row has the 4 and which has the 6, you can actually pencilmark it to prove that it forces roping.
That was a long and winding way to prove not only that the sequence exists, but it also moves the same way and causes roping in every row. (as
Proof of concept method:
Place all 5s (or 4s or 6s) to the grid in any arbitrary positions as long as you obey sudoku rules. After that surround them completely with 19 middle pencilmarks. As soon as you fill a single digit anywhere it will immediately populate all 1s and 9s including the ones that go over to the other side (where 5 is on the edge). Then those 2 rows where 5 is on the edge will force the position of 6 and 4, 2 and 8 and so forth, until the whole grid is populated simply by enforcing sudoku and dutch whisper rules.
If you go back to the point where you decided between a 1 and a 9, you can reverse the decision by using the other digit, and the grid will again fully populate but just in reverse order.
You don't need to prove anything to solve the puzzle, in fact proving things makes the puzzle much more basic and less satisfying.
@@HP4L16 Making a move without proving it is just bifurcating.
@@MarcusTheDorkus No it isn't, you can solve the puzzle without assuming anything using only the given rules and sudoku. In fact trying to "prove" it using trial and error, which is what most people who "proved it" did is actual bifurcating.
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.
While an individual line might work with another order, I suspect you can show that doing so makes some other row not work.
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.
I encourage you to keep trying, Trust me there is a logical way to prove that there is only one possible sequence that works and the only rules you need to prove that are dutch whispers and normal sudoku rules. If you need a hint then I would say that all 9 rows are needed to prove it (the secret is in the columns).
I'm not embarrassed to say I was only able to finish it using a spreadsheet.
Brilliant and very enjoyable puzzle.
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.
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.
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.
I solved from the bottom right corner, deducing the direction of the sequence from the grey square and < interactions. From there, the middle 3 rows can be deduced in the same way. I ended up with the bottom 6 rows done and then had no idea what the skyscraper rules meant and had to watch the video to understand. The rules are not well written.
14:11
Nice and simple, but cute and elegant.
The whisper lines double (and triple) as region sum lines, and modular lines.
I didn't think of the modular lines. But for the second option: If you mix a little red colour into your blue region-sum-lines, then they'll become what they really are.
Definitely not region sum line, 5+6+1=2+3+7=/=4+8+9
They are modular and renban
Definitely not region sum lines
6+5+1=2+3+7=/=4+8+9
They are modular and renban
31:25 A mathematicians dream is my nightmare lol
Its also fascinating that the digits kind of corkscrew in ascending order through the three sets of three rows as well
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.
At 14:36, Mark missed that there is a a set of valid row beginning with 62 and ending with 15: 6 2 [789] [34] [89] [34] [789] 1 5. These rows do not wrap, of course, but you must prove why they do not work anywhere.
28:50 for me. Not easy to prove a unique solution to this puzzle, but I think I followed a logical solvepath. Interesting one.
14:31. I first figured that row 3 had to have 4 and 8 at the ends to get a double 2 sky scraper, giving me 6 in the middle. Row two needed to have 2 and 7 at the ends to get double 3, giving me 9 in the middle. Then I figured the only way to make the arrow work was to make it 1+2=3, which gave me the direction. Final I figured the order of the bottom three rows with the and that the square had to be even and > 2
One of the best puzzles I have managed to solve. Very satisfying logic.
Finished in 45:41. Interesting way the Dutch lines interact with each other and particularly the skyscraper clues. It was easy to make a bad assumption, but since the places are so limited, it makes it rather simple to backtrace the errors.
Fun puzzle!
Worked it out entirely in a notepad app beforehand. Then just filled it in. Wrote out the only sequence that works, just had to figure out which direction it went and where it started on each row. For the top 3 rows, I figured out where each sequence started in each row, but not the direction. That told me that each row among those 3 must shift by 1 for another set of boxes and another 1 for the final set of boxes. There was only 1 option within those that worked for boxes 7, 8, and 9, which told me the direction (for every row) and also which set was in each row in the bottom. That gave me rows 1, 2, 3, 7, 8, and 9. For the middle row, it was pretty simple to find the only possibility for the order of the rows using the arrow clue. Whole thing took 14:51.
I sort of caught onto where he was going with this by miracle standards and just solved based on miracle. They are Dutch modular renbans.
I usually don't read comments much before leaving one myself, but this time, because you invited folks to give other logical approaches, I was curious to see what folks said. Maybe there are others? I can see that there may be other sequences. But I don't think that anyone has proven either your sequence to be the right one nor that there are not more than one - but I am not really reading carefully at this point, just skimming to satisfy my curiosity in a general way. Nonetheless, it was fascinating to hear you reason out how it all might work, and how you came to the correct solution. I will not be trying this puzzle myself, but I am very glad to have had the chance to watch you do it. By the way, "torus" and "toroidal" are words I had never heard before watching this channel. An education on so many levels. Thanks, Mark, as always.
I loved that. Had to think a bit to get started in box 9. Then I decided that the 2 rules rule would have to be significant. Renban was obviously 1 rule and then some sort of cycling. Quickly that was obviously modular. I was slow to notice that each column in a box was strictly 1 modulus. Thereafter it fell out very quickly.
they are also renband lines
Lucky 6:52 solve, starting from the 2-2 skyscraper line with the number sequence in the correct direction.
Not really a 'solve' if you just managed to find a way (lucky or not) that filled in the grid in a valid way. Why? Because you have not demonstrated that your biffed solution is unique. Still - if you are happy to not work through the logic - and assume uniqueness - then who am I to challenge? ;-)
@@schubertuk Actually I'm taking it the opposite way around: If a puzzles solution is not unique, then, depending on the way you see it, either it is not solvable (thus an unfair win to an unfair task is fair game) or any solution counts. Vice versa, IF the puzzle must solve uniquely, that is a rule in itself that can be utilized.
00:26:18 for me. Amazing puzzle! Break in was very fun to figure out. Kind comment.
An easy proof of why the the rows of dutch line must be cyclic (Open the puzzle for easier understanding)
(we say 123456789 and 912345678 and 891234567 are cyclic)
Things to know:
1. 5 can only be next to 19, 4 next to 89, 6 next to 12 on the dutch line
2. After we put in 5, everything else alternates between big (6789) and small (1234)
Preparation: Put all 5's in the diagonal, r1c1,r2c2,r3c3... We will ignore the boxes in this proof. Just look at the rows and columns
Step 1: What will happen when 5 is in column 2?
Assume it is 159xxxxxx, as everything alternates. It will be 159sBsBsB, notice that 6 must be next to 1 or 2, so it must be at the end as 26. 4 must be next to 89 so it will be 948. Result: 159483726 is the only sequence.
(If we start with 951, it will be 951627384, we will now refrain from talking about this second possibility, as it is just swapping 1-9,2-8,3-7,4-6)
Step 2: What will happen when 5 is in column 4,6,8?
Notice that row2 is now 159483726, column 3 cannot have another 9, so row4 must be xx159xxxx, continue with this logic. row 2,4,6,8 will all have 159, not 951
Step 3: We can now fill in everything in row 2468, and it will be cyclic. Hint: row 8 can be filled in a way similar to step2.
(159483726, 261594837, 372615948, 483726159)
Step 4: 5 in column 3,5,7
Notice now column 1 already have 1234 (small), so if column 2 is big, column 1 has nothing to be placed. It cannot be x951xxxxx, it must be x159xxxxx.
Repeat the process in Step 2, row 3,5,7 will all have 159. Now fill in all the 1's and 9's (row 1 will be 59xxxxxx1, row 9 will be 9xxxxxx15)
Step 5: Fill in everything else. (hint: fill in row 19 first, then look at column 4,6) Remember the two things on a dutch line!
Conclusion, All rows on the grid must be cyclic
Has anyone proven that Mark's is the one and only solution? Seems like a huge shortcut was taken in dismissing non cyclical rows which happens to work because that's what the author intended.
I'm irrationally annoyed by how they've got the usual "These Two Cells Add To Five" V symbol there instead of the properly wide inequality symbol to mean "R8C9 is bigger than R9C9".
That actually made me go back to hitting the rules again, whether I missed the five symbol in the rules.
Not irrational at all.
The caret doesn't bother you though?
@@colej.banning2419I’d never heard of caret, everyday’s a school day.
If you use an inequality sign, which cell is the bigger between C9 R7, R8 and R9?
Super fun puzzle!
8:38 here, once you figure out the cyclic order, doing the arrow sum in the middle solves the puzzle.
Oeh I’m gonna try this one! Love your videos by the way, it’s weirdly relaxing and intellectually spicy at the same time
This is a stratospherically smart and cool constrution.
👏👏👏👏👏👏👏👏
I’m seeing a bit of discussion about how to prove the cyclic sequence so I decided to try it on my own and this is what I came up with (I did see one comment by RecreationalCynic which spurred my thought process so the credit for determining that the cycles all go in the same direction is theirs and not mine)
First, think about the 5 in all of the even columns and assume that the cycle goes 159 for the 5 in c2, the 9 being placed in c3 means that the 5 in c4 must have a 1 to its left in c3 which means that the order of the sequence in one even column will force the same order in at least one other even column 5 which will then propagate and force the order 159 in all rows where the five is in an even column (this was the point where I stopped reading their comment and decided to approach it for myself, so I don’t actually know how they proved the order must be the same for all the 5s in an odd column)
Next, consider the 9 in c1, because we have already placed a 5 in c2 there can’t be another 5 in that column meaning 5 only has one neighbor (1), and must go in c9 with a 1 in c8, this then forces the order in a row where the 5 is in an odd column so the effect again propagates and now the same 159 order gets forced in all rows where the five is an odd column, and the same order being forced wherever there’s a 5 in any even or odd column just means that it’s forced in every column
The next digit to consider is the 6 in the row where the 1 is in c1, because it can’t go next to the 1 it only has one neighbor (2) placing it in c9 and being that it’s the only 6 that can be there every other 6 goes next to a 1, this can then place all of the 2s next to them
The same logic can be used to place a 4 in c1 where the 9 is in c9, which can then place all of the 4s and then 8s
Then the 3s can be placed next to the 8s and the 7s next to the 2s
This completely ignores row order, but it does prove that every row must be that same sequence just shifted
So close to pairwise solution but the cluster of rules in B9 obviously breaks the duality...
Did not think pairwise could exist w arrow sum but there existed a 2nd sum. If the C9R89 < did not wxist could probably form a pairwise.
Fun puzzle!
That arrow was easier to solve if you use already written 3 rows and search there possible location for that arrow and then you see that diagonally these digits are consecutive so that reveals these must be 1 2 3 there as using 7+8 = 9 is nonsensense. Or 8+9=1
Once you understand the way the grid needs to be laid out, actually filling in the digits is easy. I got an 8:26 on this puzzle, but that doesn’t include the 10+ minutes I spent figuring that out.
Interesting one, I don't think I ever fully grasped this one, despite solving it, and I didn't really follow Mark's reasoning.
20 mins for me. Thanks.
49:29, spent like 5 minutes not seeing how to go, then immediately had an idea on my walk and started pencil marking when I got back home. My solve was basically going "if this was a 9" in rows 1, 2, and 3, pencil marking everything, then reversing all that work in case it was the other direction. Managed to solve from there, but it was little logic, lot of trial and error.
11:05 solution 957 Roping was my friend....
I get the bonus points -- renban is obvious, and the other is modular (mod 3 rules). Vertically, each column was entropic in that each block had one from low, medium, and high. Also vertically, each block had the same modularity. Roping was everywhere, both vertically and horizontally. Does this make up for me taking forever to do it?
I noticed that Simon says debut as "DEB-you", and Mark says "DAYB-you". Is that a regional UK thing? Are they not from the same area?
_"It's sort of a prove by example"_ (Mark @13:50)
Unless you provide a method to generalize, that prove is only valid for that particular example.
I used the other method, as explained by *Tomato Pie.*
I doubt your example can be generalized, as it does not consider interactions with other rows. I doubt this is enough to prove *cyclicity.*
Well, one of them is Renban, obviously. The other one? Hmmm… 🤔
Something involving Germs?
It would seem “the lines must loop” is more of an assumption than a deduction. It might be true, but this solve really left me unsatisfied.
The Aptly named Nontuple
I dont get why mark didnt pencil make the even square bottom right as 48 sooner?
This was the first one i solved faster then here, got it in 10 mins and im by no means clever, this one just intuitively clicked
I enjoyed solving the first three rows of this puzzle, but after that it got a bit frustrating. I could see that it was going to be the same cyclic sequence in every row - probably with roping - but couldn't prove it. I ended up writing out the full contents of the remaining six rows and then rearranging them until the other clues were satisfied. I would love to know if anyone has found a more elegant route to the solution.
Things to know:
1. 5 can only be next to 19, 4 next to 89, 6 next to 12 on the dutch line
2. After we put in 5, everything else alternates between big (6789) and small (1234)
Preparation: Put all 5's in the diagonal, r1c1,r2c2,r3c3... We will ignore the boxes in this proof. Just look at the rows and columns
Step 1: What will happen when 5 is in column 2?
Assume it is 159xxxxxx, as everything alternates. It will be 159sBsBsB, notice that 6 must be next to 1 or 2, so it must be at the end as 26. 4 must be next to 89 so it will be 948. Result: 159483726 is the only sequence.
(If we start with 951, it will be 951627384, we will now refrain from talking about this second possibility, as it is just swapping 1-9,2-8,3-7,4-6)
Step 2: What will happen when 5 is in column 4,6,8?
Notice that row2 is now 159483726, column 3 cannot have another 9, so row4 must be xx159xxxx, continue with this logic. row 2,4,6,8 will all have 159, not 951
Step 3: We can now fill in everything in row 2468, and it will be cyclic. Hint: row 8 can be filled in a way similar to step2.
(159483726, 261594837, 372615948, 483726159)
Step 4: 5 in column 3,5,7
Notice now column 1 already have 1234 (small), so if column 2 is big, column 1 has nothing to be placed. It cannot be x951xxxxx, it must be x159xxxxx.
Repeat the process in Step 2, row 3,5,7 will all have 159. Now fill in all the 1's and 9's (row 1 will be 59xxxxxx1, row 9 will be 9xxxxxx15)
Step 5: Fill in everything else. (hint: fill in row 19 first, then look at column 4,6) Remember the two things on a dutch line!
All rows on the grid must be cyclic
All the orange bands are also technically purple renbans, since they're all 1-9.
I feel like I flew threw that at 19:11; I seldom complete puzzles shorter than the length of the video.
It's simply a matter of determining where exactly the forced roping begins!!! Solved in 7m13s.
So I found a solution that worked but it wasn’t the answer 😬
What is it? I'm almost certain there isn't one. - Tomato Pie
14:52 for me
I'm confused, what's the point of the skyscrapers?
I think that it has been used to fix which rotation of the sequence must be in each row, based on how many digits can be "seen" from each side of the puzzle. Without them I am not sure how one would determine that. At least, that is how Mark used them.
Without the skyscraper clues, you could take Mark's solution and swap rows 1,2,3 and obtain alternative solutions. The skyscraper clues force uniqueness.
Rules: 02:31
Let's Get Cracking: 04:59
And how about this video's Simarkisms?!
Sorry: 7x (09:31, 09:44, 10:19, 12:44, 17:10, 22:09, 29:32)
Obviously: 3x (11:13, 29:23, 33:07)
Bother: 2x (20:46, 23:29)
Fascinating: 2x (01:01, 04:41)
Goodness: 1x (30:53)
Nonsense: 1x (32:28)
Clever: 1x (27:46)
Approachable: 1x (01:54)
Hang On: 1x (25:29)
Surely: 1x (32:31)
Proof: 1x (14:32)
In Fact: 1x (13:24)
Intriguing: 1x (26:53)
Weird: 1x (27:41)
Most popular number(>9), digit and colour this video:
Ninety Four (7 mentions)
One (57 mentions)
Orange (6 mentions)
Antithesis Battles:
High (2) - Low (1)
Even (9) - Odd (0)
Outside (2) - Inside (0)
Row (41) - Column (7)
FAQ:
Q1: You missed something!
A1: That could very well be the case! Human speech can be hard to understand for computers like me! Point out the ones that I missed and maybe I'll learn!
Q2: Can you do this for another channel?
A2: I've been thinking about that and wrote some code to make that possible. Let me know which channel you think would be a good fit!
13:31 - After you have the 615948372 sequence (reversed here), it looks you can swap the 2 and 3.
On that exact sequence, yes you can, but using the sequence as a torus where is wraps around to the beginning, the 3 and 6 would be next to each other. That's a difference of 3, not 4.
@@pacman52280 Not sure it needs to be a torus, if other permutations also work?
@@pacman52280 The grid isn't a torus. Mark is trying to prove that, nonetheless, the toroidal sequence must be used in every row; and, he ends up doing so by claiming, falsely, that there are no other valid sequences for any row.
In his row 4, that has 5 in the middle (726159483), you can also swap 7 with 6 and/or 4 with 3 too.
I ended up assuming that it would be the toroidal pattern cycling and just seeing if that worked. I wasn't surprised that it did, but I knew I hadn't proved uniqueness. I was looking forward to seeing how Mark would achieve his step (B) once he mentioned it as a necessary step in the proof. Most disappointing that he didn't achieve this.
Did not like this puzzle at all. I sort of tried to play around and see "what happens if we put this here" and how it affects other things. Started in the middle and the arrow accidentally lined up and worked, so the middle 3 rows were correct, then tried to see what would satisfy the skyscraper clues and then the bottom clue and whoops - it all solved and the solution "was correct", but I had no idea why it's unique and how to prove it. Watched the video and turns out Mark just remembered if from somewhere else and sort of "saw" that it would work here. Not very satisfying - I would have liked a full apriori logic of how to deduce this from scratch and why it was necessarily the only solution.
30:29 for me