Thank you so much for solving the puzzle was really a pleasure to watch. After the set a bit easier way to find the digits which I think you would appreciate is considering as equation 3x+4y=6z and it means y is divisible by 3 which is only 9 by pencil mark and then x is even 6 or 8 but x is too high. As for the title name I was thinking about a pun and I was thinking lets do a zip code reference and searched on google that Miami has the zip code 33333 the reason was because of given 3s. Thank you again for the feature and take care!
Very smart puzzle. It was brutally hard from start to end but so fascinating that it deserved my maximum effort to solve it. Was it really so difficult to proceed with eliminations after locking the *236* triple in *box 9?* For instance, was there an easier path at 1:15:57?
Hi Dorlir. First of all, that was an incredible construction (the way the digits kept bouncing on and off the ‘U’ shaped zipper line was magic and that after a quite delightful break-in). Second, I’m interested to know what your intended process was for ruling out double 4 from the last 2 pairs on the longest line. I ended up having to look at where 4 went in the bottom row for the one and a skyscraper on 5s combined with row 7 for the other! Was that the intended path or was there a different way of proceeding? In any case, thank you for setting it (and I’m pleased that I did manage to solve it, as I was completely stuck for about half an hour).
Spectacular. I had seen this solved twice previously so I knew what the break-in was. I absolutely never would have got it myself in a million years, but it is a thing of beauty. Watching the first half an hour of the video, seeing Simon use his brilliant intuition and investigational skills, and 'setter logic', was massively entertaining to me. It was like a classic Columbo episode, where you know the answer from the beginning, and then watch in awe as he pieces together the clues. This is why I love this channel. Fantastic work from both Dorlir and Simon
Almost 2 hours for Simon? I'm not going to try, I'm just going to watch. 2 hours later: That was amazing, Simon. I had no chance of getting that. When you got to the end forgot to finish the last zipper, I can sympathize.
a quick way to reduce the options at 46:00 is to note that both green and orange are multiples of 3, so yellow does too(i.e. we have 3*(2*orange - green) = 4 yellow -> 3 is a divisor of yellow). so yellow must be 9. we can also say with the same reasoning that green must be even -> 6 or 8. the only 2 cases for orange are now (4*9+3*6)/6 = 9 or (4*9+3*8)/6=11. since orange cant be 11, we have our 3 digits. all of this logic can be done without any of the work reducing orange's possibilities which was another source of bifurcation
I am always disappointed when I have to watch a long video in more than one session, especially with such a fascinating puzzle as this one! I am not surprised that you would be tired, Simon - I can't imagine how much stamina is required to solve these longer, harder puzzles. The set stuff was amazing - I never tire of watching you figure out how to employ that technique in a sudoku puzzle. Thanks for the video!!
1:41:47 Oh is that dangerous? It might be dangerous, but let's live life on the edge. I'm seeing a t-shirt, with a mostly filled in Sudoku grid and the quote "Let's live life on the edge" :D
When you get a digit, you really should do a brief sudoku check on the column, row and box (painful as it may be!) It can do a lot to eliminate roadblocks (like that missed 2/4 with a 4 staring straight at it 😅).
This makes sense, but it is not as simple as it seems. What if your brief sudoku check allows you to get another digit, before you manage to finish it? Logic may branch in several directions and you cannot follow them all at the same time. I am sure we all choose to follow the first branches we can see. One or more, but no one can always check them all systematically. I agree that Simon sometimes scans sloppily but I am sure he would be able to do it much more accurately if he wanted. I believe he just prefers linear solving. Every time he finds a digit he likes asking the question "what does it mean"? Most of the times the answer is not found by plain sudoku, and most of the times Simon's strategy works much better than mine. In this case, for instance, I was not able to solve myself and Simon was able to find several smart tricks for eliminations that I missed completely, for instance in *column 7* and *row 8.*
I think the easy way once you got the set was to see that orange is divisible by 3, the line in box 7 is divisible by 3, so the line in box 9 must also be. I also think that's why you need to remove 6 as an option in box 9 cause that would be divisible by 3 and allow for orange and the other blue line to be 8 and that would work. So that given 3 is on of the prettiest given digits I've seen in a long time
Yup - if orange divisble by 6, then it's divisble by 3. Blue in box 7 is divisible by 3, so blue on right is divisible by 3, and by 4 because of zipper. So it's sum is divisble by 12, and therefore is 24 or 36. 24 doesn't work so you get a 9 in the corner and that travels round the grid...
I consider solving this in 54 mins an achievement. It goes a lot faster once you write the equation 6x(r4c5) = 4x(r9c9)+3x(r9c2) and reduce it modulo 3 and modulo 2, this gives the obvious r9c9 is divisible by 3, then r9c2 must be even. These immediately leave one option r9c9=9 as r9c9=6 is not possible :) Cheers Dorlir!
Glad to see this one featured, and Simon solved it brilliantly. It was very nice to see and hear Simon's thoughts while he was trying to discover the SET theory for this puzzle. I recommended this puzzle cause to me the SET was totally not expected and it opened my eyes for new possibilities regarding zippers. When I tried it , i was looking for 2 sets where i could use the U-shape zipper at the top to eliminate cells from both sets. I never expected the corner-zippers to be involved at all. But once (after a hint of Dorlir) i figured it out, i did not expect that e mathematical equation like 3x+4y=6z to be that restrictive. But it actually was the restricting part since x,y,z are obviously limited to natural numbers. And it's the beautiful nature of zipperlines that you can deduce some cells to be the equivalent of something like 4y, which wouldnt be possible with any other clue. So yeah, that's the part about the opening that really blew my mind. Well done to dorlir for setting, and obviously to simon for solving it.
1:30:07 finish. I was lucky enough to find the break-in right away, and what Simon missed at first about comparing the sets might be easier to see with an equation. I labelled r9c2 as "A" (5-6-7-8-9), r9c9 as "B" (7-8-9), and r4c5 as "C" (8-9). So the equation is 3A + 4B = 6C. If you rearrange it, you get 3*(2C-A) = 4B. This means that B has to be a multiple of 3, so it must be 9. Then it is only a question of what works to find that A=6 and C=9. Of course, then I got stalled for a good twenty minutes near the middle, trying to work out where to go. A very nice challenge, excellent!
Around 46:00 I also noticed, that you can probably reduce the numbers for the centre dot on the left blue area. Both orange and the right blue, are a multiple of an even number (6 and 4 resp.), and become even themselves if added up. Meaning the two blue regions must add up to an even number to match orange, which is only possible if the left blue also adds up to an even number, and as it's a multiple of three, it would only work if you use an even number (6 or 8). At least I hope that is right. No math pro myself. It would've probably only helped with doing the calculation in your head easier.
@@brianj959 No, both Mark and Simon are more intent on solving the puzzle and overlook the Sudoku. Simon knows this and often says that he's probably missing the Sudoku.
Easy way to see the set trick: do exactly Phistomophel/Art Van Der Wetering, except with a side box instead of a centre/corner box (ie compare columns 3-7 minus box 2 to rows 6-9). That gives an equality between the 2x4 rectangles in the bottom two corners and the region with corners in R1C3 and R5C7 minus box 2. Cancel the two 3s, and the matched pairs on the long line that intersects both. That gives an equal sum for the top square-y line and the two bottom corner lines. Also, considering parity instantly tells you that R9C2 has to be even, once you've got that, since 3 times that is equal to 4 times (something) plus 6 times (something), which is even, which then massively cuts down how many possibilities you have to check for the three lines. Repeating that mod 3 gives you a 9 in the bottom-right. I also did the whole back half of the puzzle completely differently, working out the digits on the vertical bits of the big U zipper first, then going through it all basically backwards to how Simon did it.
Simon's reasoning that led to the first digit was exactly the way I did it (but I needed to take his set approach from this video). Honestly, if I hadn't jotted the process down on a piece of notepaper I would have gone completely bobbins. My brain simply doesn't do multiple-layer problems without help. Is there something in the style guide that prevents jotting stuff down? Too analogue? Seems a rod for your own backs. Might the office expense account stretch to a small whiteboard and a pen? (I do, of course, love this channel!)
49:00 An obvious reason, yes. If you subtract 0 mod 3 from 0 mod 3, you get 0 mod 3. Thus, r9c9 is 9, as the only option giving 0 mod 3. Then your options have to be 42-36=6, 48-36=12, or 54-36=18. Only one of these is above the triangular number for 5, 15. So it must be 18, which is 6×3.
Wow..just wow Simon. Magnifecent solving from you. Such an incredible and breathtaking puzzle Dorlir!! Would love to know how you went about setting this!
It's possible to narrow down the lower-left zipper using parity. 6x = 4y + 3z, both sides of the equation must then be even, and 4y is even. Therefore 3z is even, so z (the digit in r9c2) must be a 6 or 8. That made the case checking much simpler for me.
42:45 -- As soon as you get the relationship 6x = 3 green + 4 yellow, you could've immediately deduce that yellow is divisible by 3, and knowing that it's 789 -- it becomes 9
Thank you to Simon and Dorlir for an entertaining evening. Not just for being able to solve the puzzle, and watch how someone else went about it, but for giving me a good laugh upon discovering my sarcastic, based-on-nothing break-in guess turned out to be correct.
47:19 9 is divisible by 3. So the only digit from 7, 8, 9 to make 9×6 - 4d divisible by has to be 9, because we are saying 9×6 - 4d = 3n which means 4d = 3(3×6 - n) Meaning d has 3 as a factor. Another way is to take mod 3, making d = 0 (mod 3)
Day 2 of asking for the very long solves as bonus videos Thx for the longer video today, but it s not quite long enough for some of us, we need those as we like to suffer and grow back our attention span from tiktoks and the very short content we usually see😉
What a brutal puzzle. I was so pleased when I found the break-in (which took me a lot of effort), but then I still couldn't finish it. I knew it was all about row 8, but I didn't see how to resolve it, so reluctantly I bifurcated. Well done, Simon.
The combnation of central digits can be derived looking at modulos only. Start with the big U at the top, that is divisible by 6, so you can say it is 0 mod 6. Now look at modulo 6 of the two J zippers: The small one on the left alternates between 3 mod 6 and 0 mod 6 as it's a multiple of 3. The larger one on the right will have modulo rests of 2 and 0 when you go throug the options 789. The only combination of modulo rests from left and right has to be 0 mod 6 options. That's 9 for the big J and an even number 6 or 8 for the small j in box 7. The options 24 and 36 sum to 60 though, and you can't put 10 on r4c5, the center of the big U zipper, so it must be the 6, not the 8 on the small j. There you have it with mod logic. The ingredient of maths logic to know is that when you add two numbers with a specific modulo rest for a number n, the rest mod n for the sum is the same as the sum of the rests mod n (i.e. two numbers 3 mod 6 added together have 3+3=6, but 6 then goes back to 0, so you build the modulo of the sum of the modulo rest). That makes it simpler to compute with than with the larger numbers. As the big J has no modulo rest 3 the odd numbers with modulo 3 rest can be discarded, as they would need a 3 mod 6 partner, but there are only 0 mod 6 or 2 mod 6 available. So the only valid combinations are 0 mod 6 on both sides.
29:00-41:00 was magical!! Love to see set logic in action and being the actual correct (havent continued, could turn out to be a red hering 🤔😳) way forward. When do you start to think about set logic? As in, what clues or lack thereof gives you the inkling to try this approach? Is there a general giveaway or do you count the amounts of butting your head against the logical wall? 🤭
I got a slightly different break-in, and a bit more straight forward than Simon's, which makes me so proud of myself, for once😅! I'm very happy I was able to solve this one ! Thanks for the master class Simon and Mark! It's working 🙂
It's curious there are trends everywhere, even in variant sudoku world After fog of war appearance, the new recurring variants are zipper lines and self-counting circles I assume constructors enjoy exploiting the range of possibilities each new variant may offer in terms of interesting logic
It's not that surprising, though, is it? I mean, setters are of course inspired by each other's work, so when someone makes something groundbreaking or novel, others will try to reiterate on that idea.
@ 13:39 - "That suggests that this is at least a 6" - You already pointed out in box 2 that there must be at least a 6 on that line, forcing it to be at least a 7-zipper. You also counted the cells on the line in box 4, and found there were six different digits, and also that in R5 there must be at least a 6 on the line. How could it possibly be less than a 7-zipper? At least one of R4C5 and R7C1 must be at least 8, because there's at least a 7 on one of the lines in box 4. If R7C1 is 9, by sudoku, so is R4C5. Whatever you put in R7C1, then there must either be at least a 7 in R4C3, making R4C5 at least 8, or there's an 89 pair in the empty cells in box 4, and the one in R6 must be in R4 in box 5, so that line must be at least an 8-zipper @ 25:25 - "which means that is at least 6 in the corner". No, it's at least 7 again. If it were 6, all of the digits would have to be different. Imagine if say 2 was in R6C9, that would put a 4 in R7C8, but now you'd have to put your second 2 into R8/9C8, where it must pair with a second 4 in R7/8C9. If all the digits have to be different, there's at least a 6 on the line, so it must be at least a 7-zipper. The only possibility for a repeated digit is double 4 if the line total is 8. That would push 4 onto the large U line, making it have to be a 9. @ 49:40 - "I can use the secret" - why make it hard? If it's an 8 line, what digit can't go on the line? 4. If it's a 7, what digit can't go on the line? 8. You were doing alright until you needed sudoku. You kept highlighting R4C2 which was marked 24, when there was a 4 in the column. If you'd not just slapped in the 4, but cleaned up your pencil-marks after placing each digit, you'd have fairly whizzed through. You didn't make anything like as much use of the verticals on the big U as you could have. You said "if that can't be 3 then the other side can't be 6", but you could have kept going with that. @ 1:37L28 - "How can this still be resisting?" - It isn't. It's rolled over and is asking you to tickle its tummy. It's you and your inability to do simple sudoku that's the problem. All of box 4 is done, which resolves all of R4, which resolves all of R5 and R6. When you suggested that sudoku would be what you were missing, you went straight to the least populated column. Why? Column 2 has five digits filled in, including one which resolves everything. I'm sure that break in was correct, it just seemed that the 3s and the way the other paired digits could be removed was too perfect for it not to be the break in. After that, you went to pieces, and it happened before your interruption, all because you didn't tidy up your pencil-marks. Saying "Oh there's a 4 here", when you were the one who put it there is failing to acknowledge the problem.
38:05 for me. After thinking about the amount of guesses (2) i wondered something... *What is the minimum/maximum amount of guesses (like guessing a cell starting from 1 to 9) required on a blank grid to solve it entirely?* where the rules are that in the minimum it works for the best grid scenario and the solver can guess any cell in the grid one at the time. for example if a random cell like R2C5 was a 7 the solver would have to guess 7 times until getting the correct number.
I used a slightly different set coloring than simon. First color: box 1, box3, r4, r5. Second color: C1,2,8,9. It gives the exact same pattern as Simon found once overlaping cells are removed.
Am I the only one who instantly noticed the rules are not specific enough? "Digits and equal distance from the centre of a purple line [...]" implies that the 3 in r5c5 and the digit in r3c5 sum to the digit in the middle of a line in r4c5, along with a multitude of other digits that are the same distance from the centre of a purple line. Like it's not too hard to realize they mean "digits an equal distance *along a purple line* from the centre of it must sum to that centre digit", mainly because the puzzle would likely break instantly in a million ways, besides the fact there is no other rule that would take into account the lines themselves as opposed to just the centres. But before looking too much at the board I was well prepared for quite the devious puzzle with impossibly many and difficult interactions between squares across the board :D
When in doubt, use SET. I should have taken this advice about 90 minutes earlier into the puzzle. I considered maths-based solutions earlier in the puzzle, but couldn't find it easily enough. Once I found the solution it was worth the struggle, though.
I have yet to watch the full thing, but the Miami zip codes are 33101, 33111, 33130, 33131 (among others). Curious if this has anything to do with the title/solution.
I got the break in pretty quickly and filled in its immediate consequences and then got stuck. I never found your logic at the bottom of the grid although I ground away at it for days. I did eventually find long logic ( I suppose bifurcation) that eliminated one possibility and gave me the centre bottom digit and some progress. But I still couldn't make further progress until I resolved the 4 question for the big middle zipper. After that the bottom of the grid resolved quickly & everything else did too. So my path was different than yours after the break in but I got there in the end.
Once you had the coloured lines, you had the equation 6x = 3y + 4z. Looking at this mod 3 gives 0 = z (mod 3), so z must be a multiple of 3, and hence is 9 given the possibilities. Looking at it mod 2 gives 0 = y (mod 2) so y is even, and must be 6 or 8. Thus 6x = 18 + 36 = 54, or 3x = 24 + 36 = 60. The first one is the only valid one and give (9, 6, 9) as the only valid solution.
In about 27:40 you said something about biffurcation, but to me this is not an exemple of that. If we count the features of Sven's software even the colours could be that. You are just pointing where 9 can be, and if one of this cell are wrong you can take 9 out of the first cell that you chose. About the puzzle: great puzzle, although I am always confused were we need to use the set theory and what cells we need to chose. 😅
You gotta trust your pencilmarks more. You made lots of brilliant deductions that you threw to the garbage and forced yourself to do it again half an hour later. You had corner marks on box 6 for the digit 1, it turned r4c8 into a naked 6 but you just decided to clutter the grid with markings you actively avoid.
49:00 I quickly saw that some of orange digits was even. And the sum of 4 lots of yellow was also even. So the 3 lots green also has to be even to make the maths work. So the green digit had to be even i.e. 6 or 8 Then from there it was easy to calculate
I believe the title is based on the 80's art style that is associated with the show "Miami-Vice". I may be wrong, but it's the impression I get between the pastel colored lines and the modern deco shapes the lines make.
Dorlir actually commented and said they wanted to do a pun based on zip codes. Miami has a zip code of 33333, so they went with Miami due to the given 3s
@@ghandigoots Oh wow that's pretty obscure. After sports teams were a bust, I was combing the internet for stuff involving Miami and a word that rhymes with zipper.
The big U immediately made me think Phistomefel....but around box 2 instead of 5. Working with that gave me an interesting equivalence: ALL orange with zipper (one middle dot and the central 3), ALL blue with zipper (3 middle dots and the other 3). Couldn't do anything with it, but I think it was a good start. 🙂 Set it up this way: Orange box 1 & 3. Orange row 4 & 5. Blue column 1, 2, 7 & 8. Ah....I see Simon got there later. Not sure why he didn't do it first.
i was thinking the break in was a weird version of the phistomefel ring where the rows were shuffled around. I saw the U at the top which is most of the ring, then thought maybe it was the ring where the top 3 rows were moved to the bottom. so a U + 5 middle cells of one of rows 7-9 = bottom left 2x2 in box 4 + bottom right 2x2 in box 6 and 4 cells somewhere in each of boxes 7 and 9. None of that gave me anything useful tho.
For a long while, I wondered why C1R6 and C2R6 were not greened, since they also could not be 4 (2 times 4) or 5 (due to the given 3), so they had to be 1267. At 1h10m, Simon finds it another way. But it would helped along a lot.
I love listening to Simon trying to parse out the Miami part of the name. And yes there is a team called the Miami Dolphins but no nothing there is a symbol for the team.
I would never have managed to do the first 90 minutes of this puzzle, however I think I could have done the last 15 faster than you did 😂 amazing puzzle, and great work getting it broken in
1:12:00 with 236 in box 9 you could find the naked single "5" in column 7 with its neighbour 4 on the other side. :D You was so close to see it. But well done at all, i never would be able to get that far on those hard puzzles. Great solving!
I did a sudoku yesterday on the train 4/10 difficulty, made up of four sudokus in cardinal directions, overlapping on the corners and leaving a 3x3 space in the middle (edit for clarification). I was wondering if there was a version, where set logic must be applied to the configuration to solve. Has anyone heard or seen that sort of riddle? Would be a nice challenge for me 😊 I enjoyed how the overlaps inform the columns and rows of the next quadrant, but taking set logic into that would be breathtaking probably 😁 Thank you for suggestions!
I got 408 minutes. This one took me a really long time. I still don't know the break in. I ended up slowly whittling down the centers of the purple eventually getting my first digit as the 9 in the central box on the purple. I, then, used a cool trick to figure out the digit in r1c7 to not be a 9 due to r4c4&6 being from the same set as what will be produced if a 9 is put into r1c7. This limits where those digits can go and either end up with too many in box 4 or nowhere to put the 9 if put into r5c8&9. That made r1c7 not a 9 and therefore an 8. After that, I slowly whittled down more numbers and bounced around the grid finally figuring out that r2c9 couldn't be a 9. It was quickly solved after that. What a rough one. I can't really imagine what the break in is, so I'm excited to see in the video.
I got next to nowhere without referring back to the video to see what happened, even the ending was baffling here, this was just super hard. The one little deduction I made before giving in and checking the video was that you could confirm fairly early on that r4c5 had to be strictly greater than r7c1. The digit in r7c1 has three possible placements in box 4: r4c3 is on the U-line, and r5c2 forces it to go on the U-line in box 1, both of which confirm the hypothesis. So the last possibility is r6c3, which places the digit in row 4 of box 5. That means r4c5 either *is* r7c1, or is greater than it. But if you try to make them the same digit, then you have a problem with r5c7, which you can prove if you play with colouring a bit must appear in the pair on r6c1-2, and therefore also appears in row 4 of box 5. If you try to put r7c1 in r4c5, then r5c7 is adjacent to it on the line, but r5c4 is not, and so the maths will be incorrect on the U-line. And that's as far as I ever got.
I might be wrong, but mustn't be there a 9 in the center of such a U-shaped zipper (always)? Because you need 4 different ways to add up to the same number (and you can't use 4+4 to make an 8). That would be my starting point, but I might have just overseen something.
In Sudoku squares, Simon did cheat, With Venn diagrams, oh so neat! Sets he'd compare, With cunning flair, And numbers popped in, complete! Nori, nori :P
I actually got the first 9 by a completely different approach by making difficult calculations but after that it was too hard and needed the SET trick. It seems there's always some SET or geometry trick when I can't solve something.
I did this puzzle alongside you and i like to see how our thought process was quite different but i finished in almost the same amount of time (like 100 minutes)
I'd like to note I messed up my pencil marks in box 2 and messed myself up right at the end by thinking I had it finished because I was missing pencil marks so I thought it could only be certain numbers after a few undos I realized my mistake and that the 4 could in fact be in the bottom middle cell of box 2
It's going to be very tough to illustrate the proverb "The darkest place is under the candle" better than just by showing this video where the tougher the logic were, the more probable it was that Simon was going to resolve it quickly 🙂
@@Ardalambdion 1.5 is my default for watching CtC after a few years, I'd hardly recognize Simon on normal speed. And for really long ones I may choose 2x, but that does require some rewinds to follow the logic.
At 13:50, instead of a bunch of unnecessary maths, you can simply deduce thar R7C1 can't be less than a 7 because there is no room for a 6, 7, 8, and 9 not on the zipper line in box 4. (Edit: he eventually figures this out)
I'm first subscriber to Dorlir Ahmeti's channel... so how long before we can get her to the 1,000 to encourage her to make content? To-> Professor Dorlir Ahmeti,of University of Prishtina... I suggest making Puzzles for the 120+ (IQ) set... I have decades ago finished the puzzles & riddles of my youth 🥰
I could be wrong, but Simon has admitted to playing Starcraft in the past and I think it's a meme from that game, where an accidental missclick can result in you losing your army. I vaguely remember Simon mentioning that as the source once, although Google isn't verifying that for me.
The end of this solve was a bit brutal to watch, from about the point he misses the '2' in R4C2 via sudoku until the end. Lots of very simple sudoku and zipper logic not being used to easily complete the puzzle. All credit to Simon, though, for getting the break-in, which I never would have figured out.
paused the video halfway through to do some algebra, i feel like a nerd. but yeah, if u do 4x+3y=54 to find the digits in the middle of the box 7 and 9 zippers, you get that x=9 and y=6
Brilliant mind, Simon, which verifies that my own mind by relative comparison is somewhere very low in the food chain; but can I at least have some marks for understanding the solution path when it is shown?!
This was really hard for me. Gave up on it last night after almost an hour. Today I did it in 80 minutes. The break in must have been stewing in my brain over night I think, but I got completely stuck for a long time after getting most of the zips but almost nothing on the lines.
Simon missed one shortcut when he got down to the orange and blue totals. The top set is six sets, so always even. The right set is 4 sets, so always even. Thus the left set must also be even, but it is three groups of either 5,6,7,8 or 9. Only 6 or 8 can be even, so it is a little less to check the maths on…
I'm the opposite of Simon... I find a distraction quite useful at times because I get a fresh perspective when I come back to the puzzle later on. I also am pretty bad at figuring out these puzzles and Simon is very good... also an opposite. 😂
Yeah after 30 minutes I coiuldn't figure out how to break into it, but I had a very strong suspicion it was with Sets with the big U... and I felt great that I was right
![ILLSLICK - KILLSHOT REMIX [FULL VERSION]](http://i.ytimg.com/vi/RIK8RrqIAzE/mqdefault.jpg)
Thank you so much for solving the puzzle was really a pleasure to watch. After the set a bit easier way to find the digits which I think you would appreciate is considering as equation 3x+4y=6z and it means y is divisible by 3 which is only 9 by pencil mark and then x is even 6 or 8 but x is too high. As for the title name I was thinking about a pun and I was thinking lets do a zip code reference and searched on google that Miami has the zip code 33333 the reason was because of given 3s. Thank you again for the feature and take care!
Your brain power is too much for mere mortals like me!
The University of Miami's logo is famously (because of football) a big U in two colours. Kind of fitting that Simon has three coloured Us in his grid
Just again absolutely incredible from you!!
Very smart puzzle. It was brutally hard from start to end but so fascinating that it deserved my maximum effort to solve it. Was it really so difficult to proceed with eliminations after locking the *236* triple in *box 9?* For instance, was there an easier path at 1:15:57?
Hi Dorlir. First of all, that was an incredible construction (the way the digits kept bouncing on and off the ‘U’ shaped zipper line was magic and that after a quite delightful break-in). Second, I’m interested to know what your intended process was for ruling out double 4 from the last 2 pairs on the longest line. I ended up having to look at where 4 went in the bottom row for the one and a skyscraper on 5s combined with row 7 for the other! Was that the intended path or was there a different way of proceeding? In any case, thank you for setting it (and I’m pleased that I did manage to solve it, as I was completely stuck for about half an hour).
Indeed I can. If I fill the first row with nine 4s, then I have placed one correct digit.
Technically correct is the best kind of correct
Programmers everywhere crying
@@Qazqi Darn...whaddya mean I can't if/else everything?!
I have managed to place 9 correct digits, with a nicely repeating grid of 9s
malicious compliance
Simon "Let's check that I agree with myself" sounds like a Murray Walker quote "Excuse me while I interrupt myself." Thank you for another great video
Spectacular. I had seen this solved twice previously so I knew what the break-in was. I absolutely never would have got it myself in a million years, but it is a thing of beauty. Watching the first half an hour of the video, seeing Simon use his brilliant intuition and investigational skills, and 'setter logic', was massively entertaining to me. It was like a classic Columbo episode, where you know the answer from the beginning, and then watch in awe as he pieces together the clues. This is why I love this channel. Fantastic work from both Dorlir and Simon
Loved how you wrote this!!
@@davidrattner9 agree! 👍🏻
Almost 2 hours for Simon? I'm not going to try, I'm just going to watch.
2 hours later: That was amazing, Simon. I had no chance of getting that. When you got to the end forgot to finish the last zipper, I can sympathize.
yeah, i should stop trying puzzles that need 1h+ video for them, this one took me 2h40, with the video as help when i get stuck XD
There is nothing better than an evening pleased by a 1h+ video of simon
Simon: let's get rid of all those pencilmarks.
5 Seconds later: misses important Sudoku that he had already figured out but deleted... x)
You make it sound like he would notice the pencil mark had he kept it haha
I came to the comments for this. Thank goodness I wasn't the only one 😅
a quick way to reduce the options at 46:00 is to note that both green and orange are multiples of 3, so yellow does too(i.e. we have 3*(2*orange - green) = 4 yellow -> 3 is a divisor of yellow). so yellow must be 9. we can also say with the same reasoning that green must be even -> 6 or 8. the only 2 cases for orange are now (4*9+3*6)/6 = 9 or (4*9+3*8)/6=11. since orange cant be 11, we have our 3 digits. all of this logic can be done without any of the work reducing orange's possibilities which was another source of bifurcation
I am always disappointed when I have to watch a long video in more than one session, especially with such a fascinating puzzle as this one! I am not surprised that you would be tired, Simon - I can't imagine how much stamina is required to solve these longer, harder puzzles. The set stuff was amazing - I never tire of watching you figure out how to employ that technique in a sudoku puzzle. Thanks for the video!!
1:41:47 Oh is that dangerous? It might be dangerous, but let's live life on the edge.
I'm seeing a t-shirt, with a mostly filled in Sudoku grid and the quote "Let's live life on the edge" :D
13:30 Love the cautious addition result of "at least six" when there are six distinct digits on the line in box 4 :-)
"LETS LIVE LIFE ON THE EDGE" had me actually laughing out loud.
When you get a digit, you really should do a brief sudoku check on the column, row and box (painful as it may be!) It can do a lot to eliminate roadblocks (like that missed 2/4 with a 4 staring straight at it 😅).
How long does it take him this time?
We don't do Sudoku on this channel.
This makes sense, but it is not as simple as it seems. What if your brief sudoku check allows you to get another digit, before you manage to finish it? Logic may branch in several directions and you cannot follow them all at the same time. I am sure we all choose to follow the first branches we can see. One or more, but no one can always check them all systematically.
I agree that Simon sometimes scans sloppily but I am sure he would be able to do it much more accurately if he wanted. I believe he just prefers linear solving. Every time he finds a digit he likes asking the question "what does it mean"? Most of the times the answer is not found by plain sudoku, and most of the times Simon's strategy works much better than mine. In this case, for instance, I was not able to solve myself and Simon was able to find several smart tricks for eliminations that I missed completely, for instance in *column 7* and *row 8.*
Poppycock.
I think Mark has tried to teach us all that
I think the easy way once you got the set was to see that orange is divisible by 3, the line in box 7 is divisible by 3, so the line in box 9 must also be.
I also think that's why you need to remove 6 as an option in box 9 cause that would be divisible by 3 and allow for orange and the other blue line to be 8 and that would work.
So that given 3 is on of the prettiest given digits I've seen in a long time
Yup - if orange divisble by 6, then it's divisble by 3. Blue in box 7 is divisible by 3, so blue on right is divisible by 3, and by 4 because of zipper. So it's sum is divisble by 12, and therefore is 24 or 36. 24 doesn't work so you get a 9 in the corner and that travels round the grid...
Yes! Longer videos. So excited!
I believe the Miami bit refers to the fact that the University of Miami calls themselves the U, and there are three u-shaped zippers in this puzzle.
At the 1 hour mark we have a revelation that’s adds to more thought and I loved every minute of trying to see it with you!!
I consider solving this in 54 mins an achievement. It goes a lot faster once you write the equation 6x(r4c5) = 4x(r9c9)+3x(r9c2) and reduce it modulo 3 and modulo 2, this gives the obvious r9c9 is divisible by 3, then r9c2 must be even. These immediately leave one option r9c9=9 as r9c9=6 is not possible :) Cheers Dorlir!
Glad to see this one featured, and Simon solved it brilliantly. It was very nice to see and hear Simon's thoughts while he was trying to discover the SET theory for this puzzle.
I recommended this puzzle cause to me the SET was totally not expected and it opened my eyes for new possibilities regarding zippers. When I tried it , i was looking for 2 sets where i could use the U-shape zipper at the top to eliminate cells from both sets. I never expected the corner-zippers to be involved at all. But once (after a hint of Dorlir) i figured it out, i did not expect that e mathematical equation like 3x+4y=6z to be that restrictive. But it actually was the restricting part since x,y,z are obviously limited to natural numbers. And it's the beautiful nature of zipperlines that you can deduce some cells to be the equivalent of something like 4y, which wouldnt be possible with any other clue. So yeah, that's the part about the opening that really blew my mind. Well done to dorlir for setting, and obviously to simon for solving it.
1:30:07 finish. I was lucky enough to find the break-in right away, and what Simon missed at first about comparing the sets might be easier to see with an equation. I labelled r9c2 as "A" (5-6-7-8-9), r9c9 as "B" (7-8-9), and r4c5 as "C" (8-9). So the equation is 3A + 4B = 6C. If you rearrange it, you get 3*(2C-A) = 4B. This means that B has to be a multiple of 3, so it must be 9. Then it is only a question of what works to find that A=6 and C=9.
Of course, then I got stalled for a good twenty minutes near the middle, trying to work out where to go. A very nice challenge, excellent!
Didn't expect to be solving diophantine equations on a saturday
Around 46:00 I also noticed, that you can probably reduce the numbers for the centre dot on the left blue area. Both orange and the right blue, are a multiple of an even number (6 and 4 resp.), and become even themselves if added up. Meaning the two blue regions must add up to an even number to match orange, which is only possible if the left blue also adds up to an even number, and as it's a multiple of three, it would only work if you use an even number (6 or 8).
At least I hope that is right. No math pro myself.
It would've probably only helped with doing the calculation in your head easier.
I love yelling at the TV when Simon can't see the Sudoku solution.
I wonder if he does it on purpose sometimes - offering a safe space to viewers like you and me to let off steam. 😂
@@brianj959 No, both Mark and Simon are more intent on solving the puzzle and overlook the Sudoku. Simon knows this and often says that he's probably missing the Sudoku.
58:48 For me, I'm quite proud of seeing 'it' almost immediately, neat use of the two givens too
Easy way to see the set trick: do exactly Phistomophel/Art Van Der Wetering, except with a side box instead of a centre/corner box (ie compare columns 3-7 minus box 2 to rows 6-9). That gives an equality between the 2x4 rectangles in the bottom two corners and the region with corners in R1C3 and R5C7 minus box 2. Cancel the two 3s, and the matched pairs on the long line that intersects both. That gives an equal sum for the top square-y line and the two bottom corner lines.
Also, considering parity instantly tells you that R9C2 has to be even, once you've got that, since 3 times that is equal to 4 times (something) plus 6 times (something), which is even, which then massively cuts down how many possibilities you have to check for the three lines. Repeating that mod 3 gives you a 9 in the bottom-right.
I also did the whole back half of the puzzle completely differently, working out the digits on the vertical bits of the big U zipper first, then going through it all basically backwards to how Simon did it.
Simon's reasoning that led to the first digit was exactly the way I did it (but I needed to take his set approach from this video). Honestly, if I hadn't jotted the process down on a piece of notepaper I would have gone completely bobbins. My brain simply doesn't do multiple-layer problems without help.
Is there something in the style guide that prevents jotting stuff down? Too analogue? Seems a rod for your own backs. Might the office expense account stretch to a small whiteboard and a pen?
(I do, of course, love this channel!)
Woot woot! Looking forward to a 3.5hr video at some point in the future!
49:00 An obvious reason, yes. If you subtract 0 mod 3 from 0 mod 3, you get 0 mod 3. Thus, r9c9 is 9, as the only option giving 0 mod 3. Then your options have to be 42-36=6, 48-36=12, or 54-36=18. Only one of these is above the triangular number for 5, 15. So it must be 18, which is 6×3.
Wow..just wow Simon. Magnifecent solving from you. Such an incredible and breathtaking puzzle Dorlir!!
Would love to know how you went about setting this!
1:41:39 - Simon *"DANGER"* Anthony...! "Let's live life on the edge," "Danger's my middle name...!
It's possible to narrow down the lower-left zipper using parity. 6x = 4y + 3z, both sides of the equation must then be even, and 4y is even. Therefore 3z is even, so z (the digit in r9c2) must be a 6 or 8. That made the case checking much simpler for me.
42:45 -- As soon as you get the relationship 6x = 3 green + 4 yellow, you could've immediately deduce that yellow is divisible by 3, and knowing that it's 789 -- it becomes 9
Thank you to Simon and Dorlir for an entertaining evening. Not just for being able to solve the puzzle, and watch how someone else went about it, but for giving me a good laugh upon discovering my sarcastic, based-on-nothing break-in guess turned out to be correct.
47:19 9 is divisible by 3. So the only digit from 7, 8, 9 to make
9×6 - 4d
divisible by has to be 9, because we are saying
9×6 - 4d = 3n
which means
4d = 3(3×6 - n)
Meaning d has 3 as a factor. Another way is to take mod 3, making
d = 0 (mod 3)
Day 2 of asking for the very long solves as bonus videos
Thx for the longer video today, but it s not quite long enough for some of us, we need those as we like to suffer and grow back our attention span from tiktoks and the very short content we usually see😉
Really?
What a brutal puzzle. I was so pleased when I found the break-in (which took me a lot of effort), but then I still couldn't finish it. I knew it was all about row 8, but I didn't see how to resolve it, so reluctantly I bifurcated. Well done, Simon.
The combnation of central digits can be derived looking at modulos only.
Start with the big U at the top, that is divisible by 6, so you can say it is 0 mod 6.
Now look at modulo 6 of the two J zippers: The small one on the left alternates between 3 mod 6 and 0 mod 6 as it's a multiple of 3. The larger one on the right will have modulo rests of 2 and 0 when you go throug the options 789. The only combination of modulo rests from left and right has to be 0 mod 6 options. That's 9 for the big J and an even number 6 or 8 for the small j in box 7. The options 24 and 36 sum to 60 though, and you can't put 10 on r4c5, the center of the big U zipper, so it must be the 6, not the 8 on the small j.
There you have it with mod logic. The ingredient of maths logic to know is that when you add two numbers with a specific modulo rest for a number n, the rest mod n for the sum is the same as the sum of the rests mod n (i.e. two numbers 3 mod 6 added together have 3+3=6, but 6 then goes back to 0, so you build the modulo of the sum of the modulo rest). That makes it simpler to compute with than with the larger numbers.
As the big J has no modulo rest 3 the odd numbers with modulo 3 rest can be discarded, as they would need a 3 mod 6 partner, but there are only 0 mod 6 or 2 mod 6 available. So the only valid combinations are 0 mod 6 on both sides.
29:00-41:00 was magical!!
Love to see set logic in action and being the actual correct (havent continued, could turn out to be a red hering 🤔😳) way forward.
When do you start to think about set logic? As in, what clues or lack thereof gives you the inkling to try this approach? Is there a general giveaway or do you count the amounts of butting your head against the logical wall? 🤭
I got a slightly different break-in, and a bit more straight forward than Simon's, which makes me so proud of myself, for once😅! I'm very happy I was able to solve this one ! Thanks for the master class Simon and Mark! It's working 🙂
Damn, that breakin... so beautiful. I don't think I ever could have gotten it xD
46:10 If you set 9, why don't you immediately set 9 for yellow and 6 for green? 9 orange -> 54 / 9 yellow => 54 - 36 = 18 (6 green)
49:20 Ah, sorry, there it is...
I like how that emergency that interrupted the recording later turned out to have been "a cup of tea".
It's curious there are trends everywhere, even in variant sudoku world
After fog of war appearance, the new recurring variants are zipper lines and self-counting circles
I assume constructors enjoy exploiting the range of possibilities each new variant may offer in terms of interesting logic
It's not that surprising, though, is it? I mean, setters are of course inspired by each other's work, so when someone makes something groundbreaking or novel, others will try to reiterate on that idea.
@ 13:39 - "That suggests that this is at least a 6" - You already pointed out in box 2 that there must be at least a 6 on that line, forcing it to be at least a 7-zipper. You also counted the cells on the line in box 4, and found there were six different digits, and also that in R5 there must be at least a 6 on the line. How could it possibly be less than a 7-zipper? At least one of R4C5 and R7C1 must be at least 8, because there's at least a 7 on one of the lines in box 4. If R7C1 is 9, by sudoku, so is R4C5. Whatever you put in R7C1, then there must either be at least a 7 in R4C3, making R4C5 at least 8, or there's an 89 pair in the empty cells in box 4, and the one in R6 must be in R4 in box 5, so that line must be at least an 8-zipper
@ 25:25 - "which means that is at least 6 in the corner". No, it's at least 7 again. If it were 6, all of the digits would have to be different. Imagine if say 2 was in R6C9, that would put a 4 in R7C8, but now you'd have to put your second 2 into R8/9C8, where it must pair with a second 4 in R7/8C9. If all the digits have to be different, there's at least a 6 on the line, so it must be at least a 7-zipper. The only possibility for a repeated digit is double 4 if the line total is 8. That would push 4 onto the large U line, making it have to be a 9.
@ 49:40 - "I can use the secret" - why make it hard? If it's an 8 line, what digit can't go on the line? 4. If it's a 7, what digit can't go on the line? 8.
You were doing alright until you needed sudoku. You kept highlighting R4C2 which was marked 24, when there was a 4 in the column. If you'd not just slapped in the 4, but cleaned up your pencil-marks after placing each digit, you'd have fairly whizzed through. You didn't make anything like as much use of the verticals on the big U as you could have. You said "if that can't be 3 then the other side can't be 6", but you could have kept going with that.
@ 1:37L28 - "How can this still be resisting?" - It isn't. It's rolled over and is asking you to tickle its tummy. It's you and your inability to do simple sudoku that's the problem. All of box 4 is done, which resolves all of R4, which resolves all of R5 and R6. When you suggested that sudoku would be what you were missing, you went straight to the least populated column. Why? Column 2 has five digits filled in, including one which resolves everything.
I'm sure that break in was correct, it just seemed that the 3s and the way the other paired digits could be removed was too perfect for it not to be the break in. After that, you went to pieces, and it happened before your interruption, all because you didn't tidy up your pencil-marks. Saying "Oh there's a 4 here", when you were the one who put it there is failing to acknowledge the problem.
It could be that the word "Miami" represents how zipper lines work.
(a = m+i)
The crossword solver in me really likes this explanation!
That’s a great deduction! 👏🏻👏🏻👏🏻
Twice I'm returning to the outer edges of a city (5) 🙂
38:05 for me. After thinking about the amount of guesses (2) i wondered something... *What is the minimum/maximum amount of guesses (like guessing a cell starting from 1 to 9) required on a blank grid to solve it entirely?* where the rules are that in the minimum it works for the best grid scenario and the solver can guess any cell in the grid one at the time. for example if a random cell like R2C5 was a 7 the solver would have to guess 7 times until getting the correct number.
I used a slightly different set coloring than simon. First color: box 1, box3, r4, r5. Second color: C1,2,8,9. It gives the exact same pattern as Simon found once overlaping cells are removed.
Am I the only one who instantly noticed the rules are not specific enough? "Digits and equal distance from the centre of a purple line [...]" implies that the 3 in r5c5 and the digit in r3c5 sum to the digit in the middle of a line in r4c5, along with a multitude of other digits that are the same distance from the centre of a purple line.
Like it's not too hard to realize they mean "digits an equal distance *along a purple line* from the centre of it must sum to that centre digit", mainly because the puzzle would likely break instantly in a million ways, besides the fact there is no other rule that would take into account the lines themselves as opposed to just the centres. But before looking too much at the board I was well prepared for quite the devious puzzle with impossibly many and difficult interactions between squares across the board :D
When in doubt, use SET. I should have taken this advice about 90 minutes earlier into the puzzle. I considered maths-based solutions earlier in the puzzle, but couldn't find it easily enough. Once I found the solution it was worth the struggle, though.
Wow! Awesone puzzle and wonderful solving!
The 4s Simon!! The 4s!! Could’ve shaved a lot of time off, but i get it. You were thinking higher level than the basics. Nice solve sir!
I have yet to watch the full thing, but the Miami zip codes are 33101, 33111, 33130, 33131 (among others). Curious if this has anything to do with the title/solution.
I got the break in pretty quickly and filled in its immediate consequences and then got stuck. I never found your logic at the bottom of the grid although I ground away at it for days. I did eventually find long logic ( I suppose bifurcation) that eliminated one possibility and gave me the centre bottom digit and some progress. But I still couldn't make further progress until I resolved the 4 question for the big middle zipper. After that the bottom of the grid resolved quickly & everything else did too. So my path was different than yours after the break in but I got there in the end.
Once you had the coloured lines, you had the equation 6x = 3y + 4z.
Looking at this mod 3 gives 0 = z (mod 3), so z must be a multiple of 3, and hence is 9 given the possibilities.
Looking at it mod 2 gives 0 = y (mod 2) so y is even, and must be 6 or 8.
Thus 6x = 18 + 36 = 54, or 3x = 24 + 36 = 60. The first one is the only valid one and give (9, 6, 9) as the only valid solution.
In about 27:40 you said something about biffurcation, but to me this is not an exemple of that. If we count the features of Sven's software even the colours could be that.
You are just pointing where 9 can be, and if one of this cell are wrong you can take 9 out of the first cell that you chose.
About the puzzle: great puzzle, although I am always confused were we need to use the set theory and what cells we need to chose. 😅
The logo for the University of Miami is a large U split in two down the middle. Maybe that's the big U.
You gotta trust your pencilmarks more. You made lots of brilliant deductions that you threw to the garbage and forced yourself to do it again half an hour later. You had corner marks on box 6 for the digit 1, it turned r4c8 into a naked 6 but you just decided to clutter the grid with markings you actively avoid.
49:00 I quickly saw that some of orange digits was even. And the sum of 4 lots of yellow was also even. So the 3 lots green also has to be even to make the maths work. So the green digit had to be even i.e. 6 or 8
Then from there it was easy to calculate
I believe the title is based on the 80's art style that is associated with the show "Miami-Vice". I may be wrong, but it's the impression I get between the pastel colored lines and the modern deco shapes the lines make.
I think that the University of Miami logo is a U very much like the top line.
Dorlir actually commented and said they wanted to do a pun based on zip codes. Miami has a zip code of 33333, so they went with Miami due to the given 3s
@@ghandigoots Oh wow that's pretty obscure. After sports teams were a bust, I was combing the internet for stuff involving Miami and a word that rhymes with zipper.
The big U immediately made me think Phistomefel....but around box 2 instead of 5.
Working with that gave me an interesting equivalence: ALL orange with zipper (one middle dot and the central 3), ALL blue with zipper (3 middle dots and the other 3).
Couldn't do anything with it, but I think it was a good start. 🙂
Set it up this way:
Orange box 1 & 3. Orange row 4 & 5.
Blue column 1, 2, 7 & 8.
Ah....I see Simon got there later. Not sure why he didn't do it first.
i was thinking the break in was a weird version of the phistomefel ring where the rows were shuffled around. I saw the U at the top which is most of the ring, then thought maybe it was the ring where the top 3 rows were moved to the bottom. so a U + 5 middle cells of one of rows 7-9 = bottom left 2x2 in box 4 + bottom right 2x2 in box 6 and 4 cells somewhere in each of boxes 7 and 9. None of that gave me anything useful tho.
Kind of makes me wonder why we haven't seen shuffled phistomefel rings yet.
For a long while, I wondered why C1R6 and C2R6 were not greened, since they also could not be 4 (2 times 4) or 5 (due to the given 3), so they had to be 1267. At 1h10m, Simon finds it another way. But it would helped along a lot.
I love listening to Simon trying to parse out the Miami part of the name. And yes there is a team called the Miami Dolphins but no nothing there is a symbol for the team.
I would never have managed to do the first 90 minutes of this puzzle, however I think I could have done the last 15 faster than you did 😂 amazing puzzle, and great work getting it broken in
1:12:00 with 236 in box 9 you could find the naked single "5" in column 7 with its neighbour 4 on the other side. :D You was so close to see it. But well done at all, i never would be able to get that far on those hard puzzles. Great solving!
I did a sudoku yesterday on the train 4/10 difficulty, made up of four sudokus in cardinal directions, overlapping on the corners and leaving a 3x3 space in the middle (edit for clarification).
I was wondering if there was a version, where set logic must be applied to the configuration to solve.
Has anyone heard or seen that sort of riddle? Would be a nice challenge for me 😊 I enjoyed how the overlaps inform the columns and rows of the next quadrant, but taking set logic into that would be breathtaking probably 😁
Thank you for suggestions!
I got 408 minutes. This one took me a really long time. I still don't know the break in. I ended up slowly whittling down the centers of the purple eventually getting my first digit as the 9 in the central box on the purple. I, then, used a cool trick to figure out the digit in r1c7 to not be a 9 due to r4c4&6 being from the same set as what will be produced if a 9 is put into r1c7. This limits where those digits can go and either end up with too many in box 4 or nowhere to put the 9 if put into r5c8&9. That made r1c7 not a 9 and therefore an 8. After that, I slowly whittled down more numbers and bounced around the grid finally figuring out that r2c9 couldn't be a 9. It was quickly solved after that. What a rough one. I can't really imagine what the break in is, so I'm excited to see in the video.
I got next to nowhere without referring back to the video to see what happened, even the ending was baffling here, this was just super hard. The one little deduction I made before giving in and checking the video was that you could confirm fairly early on that r4c5 had to be strictly greater than r7c1. The digit in r7c1 has three possible placements in box 4: r4c3 is on the U-line, and r5c2 forces it to go on the U-line in box 1, both of which confirm the hypothesis. So the last possibility is r6c3, which places the digit in row 4 of box 5. That means r4c5 either *is* r7c1, or is greater than it. But if you try to make them the same digit, then you have a problem with r5c7, which you can prove if you play with colouring a bit must appear in the pair on r6c1-2, and therefore also appears in row 4 of box 5. If you try to put r7c1 in r4c5, then r5c7 is adjacent to it on the line, but r5c4 is not, and so the maths will be incorrect on the U-line.
And that's as far as I ever got.
I might be wrong, but mustn't be there a 9 in the center of such a U-shaped zipper (always)? Because you need 4 different ways to add up to the same number (and you can't use 4+4 to make an 8). That would be my starting point, but I might have just overseen something.
You can use the same way to make the middle digit in two directions, e.g. 3+4 = 4+3 = 7, with the 3's and 4's swapping sides.
There's actually a really elegant break-in that doesn't use SET:
Just kidding, I wanted to scare you for a second.
In Sudoku squares, Simon did cheat,
With Venn diagrams, oh so neat!
Sets he'd compare,
With cunning flair,
And numbers popped in, complete!
Nori, nori :P
At 11:24 I can relate to Simon. I know the feeling of making horrendous pencil marks out of desparation.
I actually got the first 9 by a completely different approach by making difficult calculations but after that it was too hard and needed the SET trick. It seems there's always some SET or geometry trick when I can't solve something.
1:32:38 Do SUDOKU, Simon !!!!
Wowzers on the break in.
68:01
Well worth every second. That was insanely good.
Thank you so much.
i don't usually try anything that takes Simon much longer than an hour, but had some time; took me almost three hours, but i got there!
I did this puzzle alongside you and i like to see how our thought process was quite different but i finished in almost the same amount of time (like 100 minutes)
I'd like to note I messed up my pencil marks in box 2 and messed myself up right at the end by thinking I had it finished because I was missing pencil marks so I thought it could only be certain numbers after a few undos I realized my mistake and that the 4 could in fact be in the bottom middle cell of box 2
It's going to be very tough to illustrate the proverb "The darkest place is under the candle" better than just by showing this video where the tougher the logic were, the more probable it was that Simon was going to resolve it quickly 🙂
Me: This is a dead end.
Simon: There is something going on here.
Me (two minutes later): Ohh......
Ow boy, 1h45m ... there goes my saturday evening ^^
Apologies... this was a hard one (for me at least!)
When it's a 70+ video, I often press 1,5 speed.
@@Ardalambdion 1.5 is my default for watching CtC after a few years, I'd hardly recognize Simon on normal speed. And for really long ones I may choose 2x, but that does require some rewinds to follow the logic.
At 13:50, instead of a bunch of unnecessary maths, you can simply deduce thar R7C1 can't be less than a 7 because there is no room for a 6, 7, 8, and 9 not on the zipper line in box 4. (Edit: he eventually figures this out)
I thought the same thing.
He eventually came to the same conculusion.
yes he used the same logic in box 2, then talked rubbish about box 4
I'm first subscriber to Dorlir Ahmeti's channel...
so how long before we can get her to the 1,000 to encourage her to make content?
To-> Professor Dorlir Ahmeti,of University of Prishtina... I suggest making Puzzles for the 120+ (IQ) set... I have decades ago finished the puzzles & riddles of my youth 🥰
Where does the phrase "lost my army" come from? I have googled it and can't find that bit of slang, but I keep hearing it being used.
I could be wrong, but Simon has admitted to playing Starcraft in the past and I think it's a meme from that game, where an accidental missclick can result in you losing your army. I vaguely remember Simon mentioning that as the source once, although Google isn't verifying that for me.
Wow... so impressive. Mr. Simon, I just wanted to tell you it's a real pleasure watching you work 🤝
Wonder if you'll get the reference 🤔
1:32:00 Please, please, set the 2!
1:38:32 Got it!
Simon breaks into the puzzle I stalled on after a full hour in just a few minutes and then misses the 4 and 6 in box 4 for an hour.
Rules: 05:14
Let's Get Cracking: 06:13
Simon's time: 1h37m11s
Puzzle Solved: 1:43:24
What about this video's Top Tier Simarkisms?!
The Secret: 3x (59:04, 59:11, 59:11)
Bobbins: 2x (1:00:06, 1:26:54)
Three In the Corner: 2x (1:41:39, 1:43:15)
Maverick: 1x (55:14)
And how about this video's Simarkisms?!
Sorry: 21x (09:06, 11:18, 13:53, 14:18, 19:31, 26:40, 32:19, 34:22, 45:12, 1:05:10, 1:10:39, 1:13:38, 1:17:44, 1:20:03, 1:22:07, 1:22:14, 1:29:57, 1:30:05, 1:30:20, 1:34:27, 1:38:33)
Ah: 19x (08:19, 14:52, 15:04, 18:05, 22:03, 24:07, 49:36, 49:51, 53:09, 56:03, 58:56, 58:56, 1:00:20, 1:00:20, 1:01:13, 1:09:21, 1:25:44, 1:40:15, 1:41:24)
Obviously: 16x (00:55, 03:24, 03:43, 04:12, 07:14, 08:32, 10:33, 12:04, 20:27, 39:11, 40:39, 54:14, 1:25:05, 1:25:27, 1:28:33, 1:42:47)
By Sudoku: 14x (16:36, 49:47, 56:51, 1:00:28, 1:04:20, 1:05:49, 1:10:07, 1:12:29, 1:26:11, 1:31:53, 1:32:38, 1:42:15)
Pencil Mark/mark: 12x (06:20, 11:28, 1:05:44, 1:22:38, 1:23:41, 1:23:57, 1:27:19, 1:29:46, 1:33:47, 1:35:12, 1:41:06, 1:42:41)
Weird: 11x (22:53, 23:03, 27:28, 29:42, 37:49, 45:57, 47:50, 48:29, 1:14:39, 1:32:19, 1:32:27)
Hang On: 10x (11:44, 18:00, 37:21, 45:00, 56:03, 1:04:45, 1:12:56, 1:21:02)
In Fact: 7x (22:56, 32:58, 38:18, 50:14, 50:57, 54:59, 1:35:15)
Beautiful: 6x (50:11, 50:16, 1:31:53, 1:37:56, 1:43:40, 1:43:40)
Goodness: 5x (1:00:18, 1:23:09, 1:24:34, 1:38:33, 1:38:35)
Clever: 4x (24:35, 53:31, 1:19:20, 1:43:43)
Wow: 4x (47:18, 56:45, 1:07:13, 1:10:02)
What Does This Mean?: 4x (15:36, 24:22, 1:19:57, 1:38:49)
Good Grief: 3x (37:34, 1:31:17)
Going Mad: 3x (45:09, 45:09, 45:12)
Shouting: 3x (03:34, 04:06, 1:37:33)
Cake!: 3x (03:24, 03:42, 04:10)
Useless: 2x (1:35:33, 1:35:35)
Bother: 2x (1:05:16, 1:21:25)
Nonsense: 2x (35:30, 57:58)
If I Trust my Pencil Marks: 2x (1:23:39, 1:27:18)
Magnificent: 2x (1:44:45, 1:44:48)
Surely: 2x (1:10:49, 1:14:28)
Have a Think: 2x (19:25, 48:01)
What on Earth: 1x (24:50)
Recalcitrant: 1x (1:34:47)
Naughty: 1x (1:12:13)
In the Spotlight: 1x (1:43:18)
I Have no Clue: 1x (1:20:09)
Stuck: 1x (1:34:04)
First Digit: 1x (46:00)
Gorgeous: 1x (1:32:05)
Come on Simon: 1x (1:34:19)
Think Harder: 1x (1:05:20)
We Can Do Better Than That: 1x (1:23:24)
Losing my Army: 1x (05:21)
Baffling: 1x (1:37:21)
Fabulous: 1x (00:27)
Symmetry: 1x (23:06)
Triangular Number: 1x (09:44)
Most popular number(>9), digit and colour this video:
Thirty Two (8 mentions)
One (149 mentions)
Blue (33 mentions)
Antithesis Battles:
High (4) - Low (2)
Even (9) - Odd (2)
Black (2) - White (0)
Row (17) - Column (16)
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!
An American here. Every time Simon says "at all" (which is a lot) I hear "atoll." We could start calling it The Phistomefel Atoll.
1:00:25 look look look look look
add "pentomino" to Simarkisms
The end of this solve was a bit brutal to watch, from about the point he misses the '2' in R4C2 via sudoku until the end. Lots of very simple sudoku and zipper logic not being used to easily complete the puzzle. All credit to Simon, though, for getting the break-in, which I never would have figured out.
at 1:32:00 - don't worry, Simon gets the 2 at 1:38:00
114:08 for me. JC what a puzzle. Struggled immensely, (and I found the first digit before the break in, feel really bad about that)
wow. this is absolutely the most bananas set equivalence i have ever seen, by far. got absolutely nowhere by myself haha
paused the video halfway through to do some algebra, i feel like a nerd. but yeah, if u do 4x+3y=54 to find the digits in the middle of the box 7 and 9 zippers, you get that x=9 and y=6
every other combo wont give whole numbers and anything else obviously cannot be put into a sudoku.
Brilliant mind, Simon, which verifies that my own mind by relative comparison is somewhere very low in the food chain; but can I at least have some marks for understanding the solution path when it is shown?!
93:49 for me. I must have missed a million things on this one, but I somehow managed to finish it anyway so I'm happy with that.
This was really hard for me. Gave up on it last night after almost an hour. Today I did it in 80 minutes. The break in must have been stewing in my brain over night I think, but I got completely stuck for a long time after getting most of the zips but almost nothing on the lines.
Simon missed one shortcut when he got down to the orange and blue totals. The top set is six sets, so always even. The right set is 4 sets, so always even. Thus the left set must also be even, but it is three groups of either 5,6,7,8 or 9. Only 6 or 8 can be even, so it is a little less to check the maths on…
I'm the opposite of Simon... I find a distraction quite useful at times because I get a fresh perspective when I come back to the puzzle later on. I also am pretty bad at figuring out these puzzles and Simon is very good... also an opposite. 😂
Yeah after 30 minutes I coiuldn't figure out how to break into it, but I had a very strong suspicion it was with Sets with the big U... and I felt great that I was right
"Do have a go."
Who do you think I am? A superhero?
It’s Miami zipper because of the “U” and the “M” at the top