gdc here. Thanks so much for the feature! So happy to see this puzzle solved. Of course, the title is just a Britney Spears reference with any additional meaning :) Thanks for everyone involved in testing and especially to Samu Piano, a master of the aadvanced techniques.
I thought the 4x4 in the top left, and the 5x5 in the bottom right might come in to play, as you sometimes see that SET arrangement, but I guess it wasn't.
@@Tahgtahv Since a few people already commented on this: Aad's SET is part of the intended solve path and hence the reference to phistomefel (the "inventor" of SET) in the title. But it was very impressive to see how simon bypassed that without breaking a sweat.
@@MrGrog90 I bypassed SET as well, which proves it is not difficult to find an alternative path in this case... On the contrary, it would have been impossible for me to bypass SET this morning, when I solved *Quadrangularly Sparse* by _Kennet's Dad,_ as *Mark* impressively did in the previous CTC video: *What's going on in this quad?*
I did this one. And I did recall that the puzzle referenced the aadvanced techniques. So I did SET. Very surprised to see Simon solve this and not even commenting on the 'obvious' 4x4 and 5x5 squares.
Simon saying "I hope it's not all kind of micro-deductions" and then getting excited by almost all of them is magical. :) Like when I did not want to get a sweater for christmas as kid, got one, and then I couldn't stop wearing it until it actuall fell apart...
You can get the 7 in row 4 column 4 by a set theory trick: the 4x4 square in the top-left (box 1 plus the line that r4c4 is the middle of) contains the same digits as the spade-minus-its-tips plus the 9 in r6c6, and in particular has the same sum, and that sum is (from the spade) 73. Subtract the 45 from box 1 to have the sum along that other line equal to 28, and divide by 4 to get the central digit as 7.
@@BookofAeonsI've been too conditioned by past CtC videos. I saw the Aad van de Wetering pattern at the start and spent 30 minutes trying to use it in the break-in, before I realised I was overlooking the obvious restriction on 9s! Doh!! 😂 I labelled r4c4 as X, r5c5 as Y and r6c6 as Z. Aad's SET break-in pattern gave me that 8Y+Z=4X+45. I rearranged this to get 4(2Y-X)=45-Z. So Z had to be 1 (mod 4), so 1, 5 or 9. I also got X to be 7 or 8 and Y to be 8 or 9. I then spent 20 minutes getting nowhere until it suddenly dawned on me that 9s were heavily restricted. I was so sure it was going to be the Aad van de Wetering trick, I had ignored the obvious. It did mean I got that 7 in r4c4 fairly quickly after realising 9s were the true break-in.
@@RichSmith77 gdc here. That's the first math I did when constructing the puzzle (only had two lines in the grid at that point). After that I just needed to figure out how to eliminate 9 from R5C5 to reduce these 3 options down to 1. The reason you get all the other centers before that step is because I wanted to create a sense of dissonance in the sense of "now I must be able to resolve that last center" for anyone who hasn't focused in on R4C4 on that point.
@@MrGrog90 Thank you. Nice to know I was somewhat on the right path. 🙂 By the way, is it an unwritten rule among setters that the Aad set trick can only ever use the NW-SE corners, and never the NE-SW corners? I like to imagine any setter who went against the convention would immediately be thrown out of the setters equivalent of The Magic Circle. (I also like to imagine such an organisation exists 😂).
@@MrGrog90 I gave it a unique color because I had a hunch it would take off. The first 6x6 I made had loads of positive feedback. I'm glad the community enjoys it so much!
A "quick trick" you can use to find the 7 in R4C4 is, just like the Phistomefel ring, you can use set theory to derive that. The 4x4 square at the top left must always be the same as the 5x5 square at the bottom right minus 45. Then remove grid 9 from the equation and the area on the top left should be the same as the remaining area on the bottom right. Since this remainder had all cells inside the 8 zipper and you had already derived the 9, you could get that the L zipper with it's center in R4C4 should all add to 28, which can only mean a center cell of value 7. This is handy anytime you have constraints in these areas, which is a lot more than you would expect.
It seems very strange to me to call that technique "set theory", as it really doesn't have anything to do with the theory of mathematical sets as far as I can see. It's just combinators and clever counting. Still a cool way to solve the puzzle though.
This was the very first thing I did (before even filling in the 9s): I felt like the "spade" was more or less begging for it. But I also found Simon's way delightful.
One thing that is entirely unnecessary in this puzzle but might be worth noting for future puzzles of this type-if a zipper line of length 2n+1 is entirely in a box (or indeed if arranged in any way where the non-middle cells all see each other) there must be n or more ways of making that sum in two different digits; as such, the middle cell of the line in box 3 could be pencil marked 789 right out the gate since no sudoku digit under 7 can be made in three different non-repeating pairs.
I finished in 58 minutes. The geometry of this puzzle was delightful with the way the 9s bounced around the grid, followed by the rest. Figuring out that 7 couldn't be in the center cell was very satisfying to do in my head. Great Puzzle!
Absolutley gorgeous puzzle. For 80% of the puzzle I thought "there's no way i'm going to solve this" yet the logical pathways kept revealing themselves. Very approachable puzzle, but just enough to make you feel very accomplished in solving it.
I came to this video several hours after it was first posted, watched it, and absolutely loved it. I think that I can possibly do this puzzle, so I'm putting it on my list. Not only is this a great puzzle, and a wonderful video, it had a guitar intro! Thanks, Simon, for this.
The Aad van de Wetering equalit of the 4x4 square (here upper left) and 5x5 downright would have gotten you to the 7 easier. Downright is exactly one set more, so you subtract box 9. Now the downright is exactly 7 sets of 8, plus an 8 and a 9, makes 73. Subtract box 1 (which is 45, ssshh) and end up with 28. The angle around box 1 sums up to 4 sets equaling 28, that's 7.
You can do Aads set trick to deduce the 7 in r4c4 which is cool. The arrow shaped zipper plus the 9 inn box 5 has a total sum of 73. Minus the 45 and you're left with 28 as the total sum of the zipper surrounding box 1. This forces it to be 7.
That is what i did too. Then i deduced there could be no repeats on the 7 Zipper because then there would be two repeated digits that would go together on the 8 Zipper.
A small deduction I found useful to place the 8/1 pair early along the zipper for box 7's 9: since the line adds up to 10 and includes a 1 on it, the other two digits must add up to 9. Therefore, they must repeat along the other half of the zipper, and must take up the two side spots so they don't see themselves, so the 8 in box 4 must be in the center bottom.
A more slightly general statement (which I didn't actually see until after I'd finished the puzzle) is that neither what you refer to as neither the "left" nor what you refer to as the "top" can contain a pair of digits that add up to R4C4. Because that pair of digits will get pushed into R5C6 and R6C5, which means they have to add up to 8 but they don't, they add up to R4C4.
Well, Simon, a sip from the devils cup. As others noted related to set as the phistomefel ring is one of the first things explainable by equivalent sets aka set theory. But more related to Aad v.d. Wetering for the specific 4x4=5x5-3x3 set equivalence, where the 3x3 is a complete set of 1-9 as the difference of the 5x5 and 4x4.
Always interesting to see different ways to get to a conclusion. Around 21:26 Simon concludes r4c4 is a 6 or 7 by maths. I went: r4c4 is maxed at being a 7, so 7 in box 2 must go on the 8 or 9 zipper line. If r4c4 was a 6, 6 would also have to go on 8 or 9's zipper line. However if we tried making r4c4 a 5, then we have 5, 6, and 7 on the 8 and 9 zipper line, but 5+6 = 11, which can't work on a single zipper.
theres a few nice trick you can do with aad after you get the 9s on to get a lot of the deductions simon made in other ways. what aad says is that the top 4x4 cells in any corner are the identical cells to the 5x5 in the opposite corner minus the 3x3 box. so basically the "spade" zipper line plus r6c6 - the 2 bits in box 9 has the same composition as box 1 plus the zipper centered at r4c4. that means that there's an 8 on the r5c5 zipper (since there's an 8 in box 1) and it can only go in r5c5. the sum of all the cells is 8*8+9 = 73 = 45 + sum of the r4c4 zipper. so r4c4 = 28/4=7. theres a 4 in box 1, so one of the pairs on the r5c5 zipper is a 44 pair (not too useful yet, but simon did use that fact later on after finding it a different way)
33:08 finish. I also used the 4x4 and 5x5 square in my solve (removing box 9 to make each region identical). I colored the upper left 4x4 green, and the bottom right region pink. Since there has to be at least one 8 and one 9 in the green region, you need to place them in the pink region. This means that 9 has to be either in r5c5 or r6c6. If the 9 is in r6c6, then the 8 has to be in r5c5 as the biggest number available for the zipper. It was only at this point that I remembered to look for 9s on other lines, which differentiated box 5. (A 9 in r5c5 would have been more problematic for placing the 8 in the pink region, as it would have been able to go anywhere on a 9 zipper.) Then I used math to figure out r4c4: 45 + 4x (green) = 8*8 + 9 (pink). This gave me x = 7 and I continued on. Such a fun puzzle!
A bit surprised that Simon didn't use Aad's SET trick to compare the zipper lines that make L-shapes. That was my first thought when looking at the puzzle.
I similarly saw it instantly, but it wasn’t immediately obvious how to use it. The stuff with the 8s and 9s felt like a much clearer break in. Aad would have been an easier way of getting r4c4 from there, though I rather liked how Simon did it because it felt more tied to the inner logic of the zipper lines.
It's very helpful to do so as well. One can deduce the 7 in the middle box right away (73-45)/4. With a bit more effort one can also deduce that in the two legs of the L there will be one of each of the digits from 1 to 6 without repeating.
@@felicote gdc here. Very cool to hear that you used the SET way to show that there are no repeats on the 7 zipper. It's the deduction that makes the SET path more beautiful IMO.
Beautiful puzzle. Being sick and foggy, needed Simon to show me what 7-zipper line (pun intended for IT guys) meant. From there, with some ardent wrestling the logic unfolded and unraveled. Again, beautiful puzzle. Didn't watch Simon until the end, sorry Simon. Got too late. Still thanks!
Very nice puzzle! Finished in 17:46. I saw that r4c4 had to be 7 in another way than Simon. I started as Simon by seeing that c4r123 adds to 11, so if r4c4=6 then r4c123 adds to 7. That sum is 124 and that goes with 542 (to make 6-sums) on the line in box 2. But now both 2 and 4 are on both parts of that line, and therefore 2 and 4 goes in r5c6 and r6c5. That is problem because 2+4 is not 8.
Really pleased to see Simon break in with the issue of the 1-2-4 triple in box 4 clashing with box 7. I figured the same thing out, but it took me 50 minutes to get to that point so I thought I must be missing something. It was reassuring to see Simon go through exactly the same process I did (albeit much faster). Then I read the comments and realised I *was* missing something 😀. I didn't know about the 4x4 and 5x5 set theory. I'll have to remember that one! @MrGrog90 Thanks for a brilliant puzzle. Really satisfying to solve, and good to have now learned a new bit of Sudoku theory too.
Thanks! The SET way is by far the the easiest route because it also allows you to show that there are no repeated digits on the 7 zipper if you think about the actual digits (and not just the sums) in the two areas. Once there are no repeats, the digits making up the 11 sum inbox 4 are fixed and you can get to 32:07 pretty quickly. The way to show that there are no repeats is as follows: one seven on the 7 zipper => two sevens on the 8 zipper => two ones on the 8 zipper => one one on the 7 zipper => one six on the 7 zipper => two sixes on the 8 zipper => ... . You can iterate this for a bit to see that each digit from 1 to 7 must appear exactly once on the 7 zipper and exactly twice on the 8 zipper.
At 22:22 after Simon said "if these [R4 in box 4] add up to 7, then these [C2 in box 7] add up to 7" I expected him to continue with "and that's impossible because the only way to form 7 in 3 cells is with 1/2/4, and there would be no fill for R4C2" which would prove the sum is 10 and R4C4 is 7. He figured that out about a minute later, though it took another minute to enunciate it. Actually I found R4C4 to be 7 using the set-theory result that the digits in the upper-left 4x4 square of cells are identical to the digits in the lower-right 5x5 square of cells, less one set of 1-9. If you just exclude box 9 and look at the total of these cells you have the 8 in R5C5, 7 pairs of cells each adding up to 8, and the 9 in R6C6, for a total of 73. Subtract 45 for box 1 and you get the zipper line centered on R4C4 sums to 28, and this value is 4 times the zipper sum which is thus 7.
67:04, had to undo a bit of work twice due to errors I made, not counting multiple mistypes. But it was dang fun using some logic I learned watching this channel, would not have been able to solve this a few years ago.
That was very interesting. Having first worked on the 9's, I then went to the set theory that box 1 and the 'ring' around it equal the two 'rings' around box 9 and it flowed beautifully from there. Finished in 21:46. Watched the start of the vid and realising Simon wasn't going to use set, decided to do it again without set. Took 41 minutes that way as I only entered numbers when I could 'prove' the logic. Didn't flow nearly as well either. Interested to know whether gdc set this with set theory in mind...?
Fascinating to see this one solved after trying it. I did immediately think of Phistomefel, or rather the Aad's Theorem version of it, comparing the upper-left 4x4 with the lower-right 5x5 minus box 9. You get a sum of 73 in the lower-right once you place the 9, and I used that to determine the 7 in box 5, and also to argue there had to be exactly two copies of each digit from 1-7 in the lower-right set. The 7 line in the top left would allow a pair of duplicates, but only if it was a pair that summed to 7, while the lower-right L-shaped region would allow a third pair of digits summing to 8, but no pair of digits adds both to 7 and to 8. So the 16-cell regions from that set theory argument had to have exactly one 9, one 8, and two each of every digit 1-7 in order to satisfy both the set theory and the two relevant zipper lines. The rest of the solve does seem to follow a lot of the same arguments, I just ran into them in a somewhat different order. I guess there was in fact a way around using Phistomefel that a lot of others found too, but I'm pretty pleased with having successfully used a SET argument for the first time in a solve lol, even if it took me some 66 minutes to tackle.
I also tried using Aad's SET theorem for the break-in. I could only get r4c4 down to either 7 or 8 though. Until I belatedly saw that 9s were heavily restricted by the other zipper lines.
Really fun puzzle. Used a bit of wet theory to figure out what values to go in the centre of the zipper lines in box 5. Thanks Simon for teaching me that "trick" haha
Set maybe could have been used but Aad's trick. Anyway the interactions of them clues was more like pure comedy. Absolutely loved that puzzle and im going to wait a fortnight, after that i will give the puzzle a go and see how much i can remember.
There is one foggy zipper on the channel already. The puzzle is called "Lavender in the Fog" and the video title is "THIS Actually Makes Learning Maths FUN".
Simon... it's really difficult for us when you say "I'll come back to that later" and then you take forever to return to it. Many times, you forget, and it takes forever. Meanwhile, it's all we can think about, and we have trouble enjoying what you're currently doing because that's eating at the back of our mind!
Wow, I'm on a streak that is hot! These zipper lines just hit the spot! And it's strange but it's true i get through before you... (sheepishly) ...'Cause I hit the pause button a lot. (Leaves a Beach Bears Christmas album and flies off)
18:30 You could represent this by coloring the OTHER leg of that 8 in R7C4... I mean it wouldn't directly "help" per se... :D But representation it is!
Took me 92:41. I wouldn't call this approachable but it was a fun solve, if very difficult. I found some of the necessary logic like pulling teeth, determining how the sets that total 9 had to be aligned in the first three columns wasn't easy for me to catch.
Beautiful puzzle again! :-) Simon, you mentioned the name Phistomefel. I may have missed something, but probably I haven't seen a Phistomefel puzzle on this channel for ages. Is he still active?
Is it possible to still go over all the rules even if you previously went over them? I clicked the rules bookmark after opening the video but had to go back and try to find where you discussed the zipper lines because I thought I understood it, but I wasn't 100% sure. Usually, I click on the rules bookmark, watch them, then try to solve myself.
@ 32:33 - "I'm desperately looking" - You just pencil-marked 3s in box 1, and claimed that it could go on the line with 6 in R2C1. That's not possible because the 6 in box 4 pushes 6 into C1 in box 7. Therefore both 3 and 6 are in C2 in box 1, and neither can go on the line, so they form a 63 pair. This makes R1C2 a 45, so R2C1 is also 45. You now have a 45 pair in R1, which puts a 45 pair next to the 9 in box 3. That was a treat of a puzzle. My method was initially the same, placing 9s and 8s, and the 7 in box 5. Then I realised the 1 had to be on the 10 part of the 7 line, so 1 had to be on the line in box 6, so 7 had to be on the line in box 8. Lots of similar deductions gradually filled out the lines, which left me with a host of triples. The few known digits could be used to gradually resolve everything.
This is a really cool puzzle. Though I am embarrassed as to how long it took me to notice how those two lines covered that set perfectly forcing the smaller line to be a 4 + 3 sets of 8. I'm usually better at picking out that particular set, specially when it's so telegraphed like that.
Great design....manage to get all the correct sets on the lines... But failed after 80% completed cu of sudokuing err. That ruined the solve. Mostly cuz i forgot to finish the B78 line after placing the 2nd digit that led to B9 8sum audokuing err.
I enjoyed the puzzle and the solve process. Please consider removing colors from a puzzle when they no longer add value. That would make it much easier for me to see. Thank you.
RUclips broke into this with ads every 8 minutes. That's too much. Now, I'm a baby boomer and I grew up with commercial TV, so it's not like I don't understand about ads. But on regular TV, the ads are prepared and predictable: they come between scenes of a drama or comedy, or in time-outs of a sports event, or between rounds of a competition or game show. On RUclips, they just barge in, interrupting Simon mid-sentence, mid-word, maybe even mid-phoneme. It's disruptive. A couple of years ago they'd at least put a little tag on screen, "ads in 5 seconds" (counting down), so you'd know it was coming. Now there's just an abrupt change of light level and sound level, with no warning. How many ads I get seems to be fairly random; sometimes I can exit RUclips and come back later and get a better result. I don't mind ads at the start of a video, and I don't mind a block of ads in the middle of a really long one (say, over an hour), but every 8 minutes makes the videos borderline unwatchable. Unfortunately I suspect this may be out of CTC's control. 😾
Zipper lines don't need an odd length. They're similar to palindromes. An even length line would just have the same sums as they move away from the centre point.
gdc here. The way the lines are currently worded they need to be odd length. But it's true to say there is a "natural" extension of the ruleset and there is already a puzzle on LMD using it. By natural I mean, lots of people independently came up with the exact same idea of how to generalize the lines for even lengths. But I'm currently unsure if the extension adds enough to justify the increased complexity. Would you like to see puzzles using that extension?
May I suggest that the zipper line rule be rewritten to say that “digits on a lavender ‘zipper line’ that are equidistant from that line’s center must sum to the digit in the middle of that line.” The issue with the rule as written is that it doesn’t specify that it is referencing just digits _on_ zipper lines. Once it occurs to you that it means digits on the lines, the implication is clear, and the lines would be meaningless otherwise, but it can be misleading, and it isn’t technically correct.
I always have to laugh when you're playing a zipper line puzzle because zipper line sounds exactly like "Zipperlein" which is a German word for some slight, minor pain you're having.
gdc here. Thanks so much for the feature! So happy to see this puzzle solved. Of course, the title is just a Britney Spears reference with any additional meaning :) Thanks for everyone involved in testing and especially to Samu Piano, a master of the aadvanced techniques.
I thought the 4x4 in the top left, and the 5x5 in the bottom right might come in to play, as you sometimes see that SET arrangement, but I guess it wasn't.
@@Tahgtahv Since a few people already commented on this: Aad's SET is part of the intended solve path and hence the reference to phistomefel (the "inventor" of SET) in the title. But it was very impressive to see how simon bypassed that without breaking a sweat.
@@MrGrog90 I bypassed SET as well, which proves it is not difficult to find an alternative path in this case...
On the contrary, it would have been impossible for me to bypass SET this morning, when I solved *Quadrangularly Sparse* by _Kennet's Dad,_ as *Mark* impressively did in the previous CTC video: *What's going on in this quad?*
I did this one. And I did recall that the puzzle referenced the aadvanced techniques. So I did SET. Very surprised to see Simon solve this and not even commenting on the 'obvious' 4x4 and 5x5 squares.
I have no idea what SET is. I’m just happy I can solve these through sheer logic without knowing the theory. Great puzzle, loads of fun.
The lengths that Simon will go to in order to avoid doing sudoku in a sudoku puzzle is simultaneously astonishing and absolutely beautiful 😂
That's what keeps bringing me back. Out of all of the Soduko videos out there, Simon's are the most entertaining to watch.
He really does leverage all of the non-scanning-related logic to the hilt. It is uncommonly fun
Simon saying "I hope it's not all kind of micro-deductions" and then getting excited by almost all of them is magical. :) Like when I did not want to get a sweater for christmas as kid, got one, and then I couldn't stop wearing it until it actuall fell apart...
Anyone else think the CtC Patreon is one of the best values in digital gaming?
You can get the 7 in row 4 column 4 by a set theory trick: the 4x4 square in the top-left (box 1 plus the line that r4c4 is the middle of) contains the same digits as the spade-minus-its-tips plus the 9 in r6c6, and in particular has the same sum, and that sum is (from the spade) 73. Subtract the 45 from box 1 to have the sum along that other line equal to 28, and divide by 4 to get the central digit as 7.
This is the Aad van der Wetering set.
@@BookofAeonsI've been too conditioned by past CtC videos.
I saw the Aad van de Wetering pattern at the start and spent 30 minutes trying to use it in the break-in, before I realised I was overlooking the obvious restriction on 9s! Doh!! 😂
I labelled r4c4 as X, r5c5 as Y and r6c6 as Z. Aad's SET break-in pattern gave me that 8Y+Z=4X+45. I rearranged this to get 4(2Y-X)=45-Z. So Z had to be 1 (mod 4), so 1, 5 or 9. I also got X to be 7 or 8 and Y to be 8 or 9.
I then spent 20 minutes getting nowhere until it suddenly dawned on me that 9s were heavily restricted. I was so sure it was going to be the Aad van de Wetering trick, I had ignored the obvious.
It did mean I got that 7 in r4c4 fairly quickly after realising 9s were the true break-in.
@@RichSmith77 gdc here. That's the first math I did when constructing the puzzle (only had two lines in the grid at that point). After that I just needed to figure out how to eliminate 9 from R5C5 to reduce these 3 options down to 1. The reason you get all the other centers before that step is because I wanted to create a sense of dissonance in the sense of "now I must be able to resolve that last center" for anyone who hasn't focused in on R4C4 on that point.
@@MrGrog90 Thank you. Nice to know I was somewhat on the right path. 🙂
By the way, is it an unwritten rule among setters that the Aad set trick can only ever use the NW-SE corners, and never the NE-SW corners? I like to imagine any setter who went against the convention would immediately be thrown out of the setters equivalent of The Magic Circle. (I also like to imagine such an organisation exists 😂).
The circle doesn't allow me to answer that question 😂.
Absolutely splendid guitar playing from you Simon!!!
Love seeing these Zipper Lines spread!
Hey Jaze. I know I sound like a broken record at that point but thanks so much for that constraint!
@@MrGrog90 I gave it a unique color because I had a hunch it would take off. The first 6x6 I made had loads of positive feedback. I'm glad the community enjoys it so much!
A "quick trick" you can use to find the 7 in R4C4 is, just like the Phistomefel ring, you can use set theory to derive that. The 4x4 square at the top left must always be the same as the 5x5 square at the bottom right minus 45. Then remove grid 9 from the equation and the area on the top left should be the same as the remaining area on the bottom right. Since this remainder had all cells inside the 8 zipper and you had already derived the 9, you could get that the L zipper with it's center in R4C4 should all add to 28, which can only mean a center cell of value 7. This is handy anytime you have constraints in these areas, which is a lot more than you would expect.
I used that method also.
This is how I did it as well and seems to be the intended solve path.
It seems very strange to me to call that technique "set theory", as it really doesn't have anything to do with the theory of mathematical sets as far as I can see. It's just combinators and clever counting. Still a cool way to solve the puzzle though.
Yes, I noticed the Aad Van De Wetering trick as well.
This was the very first thing I did (before even filling in the 9s): I felt like the "spade" was more or less begging for it. But I also found Simon's way delightful.
One thing that is entirely unnecessary in this puzzle but might be worth noting for future puzzles of this type-if a zipper line of length 2n+1 is entirely in a box (or indeed if arranged in any way where the non-middle cells all see each other) there must be n or more ways of making that sum in two different digits; as such, the middle cell of the line in box 3 could be pencil marked 789 right out the gate since no sudoku digit under 7 can be made in three different non-repeating pairs.
Alternatively, there have to be six digits on the line all lower than the circled digit.
I finished in 58 minutes. The geometry of this puzzle was delightful with the way the 9s bounced around the grid, followed by the rest. Figuring out that 7 couldn't be in the center cell was very satisfying to do in my head. Great Puzzle!
Absolutley gorgeous puzzle. For 80% of the puzzle I thought "there's no way i'm going to solve this" yet the logical pathways kept revealing themselves. Very approachable puzzle, but just enough to make you feel very accomplished in solving it.
I came to this video several hours after it was first posted, watched it, and absolutely loved it. I think that I can possibly do this puzzle, so I'm putting it on my list. Not only is this a great puzzle, and a wonderful video, it had a guitar intro! Thanks, Simon, for this.
The Aad van de Wetering equalit of the 4x4 square (here upper left) and 5x5 downright would have gotten you to the 7 easier.
Downright is exactly one set more, so you subtract box 9.
Now the downright is exactly 7 sets of 8, plus an 8 and a 9, makes 73.
Subtract box 1 (which is 45, ssshh) and end up with 28.
The angle around box 1 sums up to 4 sets equaling 28, that's 7.
Is there an explanation of the Aad van de Wetering equalit in a CTC or other video? I would like to learn how it works.
You can do Aads set trick to deduce the 7 in r4c4 which is cool.
The arrow shaped zipper plus the 9 inn box 5 has a total sum of 73. Minus the 45 and you're left with 28 as the total sum of the zipper surrounding box 1. This forces it to be 7.
I was expecting that: Simon is such a fan of SET that I figured he'd look for this right away, though it turned out he didn't need it. 😺
That is what i did too.
Then i deduced there could be no repeats on the 7 Zipper because then there would be two repeated digits that would go together on the 8 Zipper.
A small deduction I found useful to place the 8/1 pair early along the zipper for box 7's 9: since the line adds up to 10 and includes a 1 on it, the other two digits must add up to 9. Therefore, they must repeat along the other half of the zipper, and must take up the two side spots so they don't see themselves, so the 8 in box 4 must be in the center bottom.
24:10 i got the 7 by seeing that with 124 on the left, we'd need 245 on the top, which pushes 24 on the line in box 5, and they don't add to 8 :)
A more slightly general statement (which I didn't actually see until after I'd finished the puzzle) is that neither what you refer to as neither the "left" nor what you refer to as the "top" can contain a pair of digits that add up to R4C4. Because that pair of digits will get pushed into R5C6 and R6C5, which means they have to add up to 8 but they don't, they add up to R4C4.
Omg you play the guitar its so beautiful and wholesome❤
Looking forward to zipper line sudokus in the line sudoku pack! 🎉🎉🎉
A difficult solve but very satisfying once you get rolling. Some really beautiful logic, I'm happy I was able to finish.
Well, Simon, a sip from the devils cup.
As others noted related to set as the phistomefel ring is one of the first things explainable by equivalent sets aka set theory. But more related to Aad v.d. Wetering for the specific 4x4=5x5-3x3 set equivalence, where the 3x3 is a complete set of 1-9 as the difference of the 5x5 and 4x4.
Very hard for me! Great solve by Simon who is in his element doing these puzzles!
Always interesting to see different ways to get to a conclusion. Around 21:26 Simon concludes r4c4 is a 6 or 7 by maths. I went: r4c4 is maxed at being a 7, so 7 in box 2 must go on the 8 or 9 zipper line. If r4c4 was a 6, 6 would also have to go on 8 or 9's zipper line. However if we tried making r4c4 a 5, then we have 5, 6, and 7 on the 8 and 9 zipper line, but 5+6 = 11, which can't work on a single zipper.
I love puzzles with zippers and guitar intros
Amazing puzzle. Had a lot of fun solving it. Thanks!
Simon's creative use of math is extraordinary.
A really fun puzzle. Thanks for sharing it
theres a few nice trick you can do with aad after you get the 9s on to get a lot of the deductions simon made in other ways. what aad says is that the top 4x4 cells in any corner are the identical cells to the 5x5 in the opposite corner minus the 3x3 box. so basically the "spade" zipper line plus r6c6 - the 2 bits in box 9 has the same composition as box 1 plus the zipper centered at r4c4. that means that there's an 8 on the r5c5 zipper (since there's an 8 in box 1) and it can only go in r5c5. the sum of all the cells is 8*8+9 = 73 = 45 + sum of the r4c4 zipper. so r4c4 = 28/4=7. theres a 4 in box 1, so one of the pairs on the r5c5 zipper is a 44 pair (not too useful yet, but simon did use that fact later on after finding it a different way)
Yay!!! Another guitar intro!!! 🎸☺️
Edit: Thank you very much for the heart ❤
Wonderful and pretty puzzle.
33:08 finish. I also used the 4x4 and 5x5 square in my solve (removing box 9 to make each region identical). I colored the upper left 4x4 green, and the bottom right region pink. Since there has to be at least one 8 and one 9 in the green region, you need to place them in the pink region. This means that 9 has to be either in r5c5 or r6c6. If the 9 is in r6c6, then the 8 has to be in r5c5 as the biggest number available for the zipper. It was only at this point that I remembered to look for 9s on other lines, which differentiated box 5. (A 9 in r5c5 would have been more problematic for placing the 8 in the pink region, as it would have been able to go anywhere on a 9 zipper.) Then I used math to figure out r4c4: 45 + 4x (green) = 8*8 + 9 (pink). This gave me x = 7 and I continued on. Such a fun puzzle!
Always love when I get to play Cross Sums (called Kakoru or something like that these days) while playing Sudoku.
This new rule is so GOOD!!! I like it very much!!
Simon playing the guitar is adorable! Haha, your child must like it when you play for her! 🐣
brasil
Rules: 06:33
Let's Get Cracking: 07:00
What about this video's Top Tier Simarkisms?!
The Secret: 7x (07:40, 07:44, 07:46, 14:12, 14:15, 14:24, 17:01)
Phistomefel: 3x (00:50, 00:53, 40:58)
Bobbins: 1x (08:17)
Three In the Corner: 1x (38:50)
And how about this video's Simarkisms?!
Ah: 11x (07:17, 10:06, 10:42, 11:16, 15:53, 22:15, 25:04, 29:14, 36:52, 38:42, 39:23)
By Sudoku: 9x (12:00, 12:13, 18:07, 18:10, 29:42, 29:58, 36:10, 36:34, 36:38)
Beautiful: 8x (02:13, 05:15, 23:14, 23:14, 24:10, 26:58, 34:08, 34:12)
Brilliant: 6x (06:03, 06:22, 29:12, 40:42, 40:45, 40:49)
Obviously: 6x (08:04, 13:40, 19:10, 19:33, 19:36, 20:27)
Hang On: 5x (15:51, 27:10, 35:28, 35:49, 37:12)
Sorry: 4x (05:28, 13:56, 15:57, 18:51)
Pencil Mark/mark: 3x (09:52, 12:19, 35:46)
Good Grief: 2x (04:50, 31:35)
The Answer is: 2x (10:51, 18:05)
Bizarre: 2x (23:07, 27:05)
Wow: 2x (28:49, 28:49)
Lunacy: 2x (26:21, 26:26)
Cake!: 2x (04:50, 06:21)
Goodness: 1x (40:25)
Bother: 1x (12:32)
Out of Nowhere: 1x (34:00)
Horrible Feeling: 1x (38:06)
Lovely: 1x (06:19)
Incredible: 1x (29:09)
Take a Bow: 1x (41:09)
Shouting: 1x (05:35)
Magnificent: 1x (11:48)
In Fact: 1x (14:12)
Progress: 1x (30:49)
What Does This Mean?: 1x (30:10)
Symmetry: 1x (11:21)
Weird: 1x (21:55)
Most popular number(>9), digit and colour this video:
Twenty Seven (15 mentions)
Eight (86 mentions)
Green (4 mentions)
Antithesis Battles:
Even (4) - Odd (0)
Column (7) - Row (5)
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!
1:15:37 - Phew! It took me ages to get the break in, and even then it wasn’t too straightforward. Very happy to have finished it though.
A bit surprised that Simon didn't use Aad's SET trick to compare the zipper lines that make L-shapes. That was my first thought when looking at the puzzle.
I similarly saw it instantly, but it wasn’t immediately obvious how to use it. The stuff with the 8s and 9s felt like a much clearer break in. Aad would have been an easier way of getting r4c4 from there, though I rather liked how Simon did it because it felt more tied to the inner logic of the zipper lines.
It's very helpful to do so as well. One can deduce the 7 in the middle box right away (73-45)/4. With a bit more effort one can also deduce that in the two legs of the L there will be one of each of the digits from 1 to 6 without repeating.
@@felicote gdc here. Very cool to hear that you used the SET way to show that there are no repeats on the 7 zipper. It's the deduction that makes the SET path more beautiful IMO.
Beautiful puzzle. Being sick and foggy, needed Simon to show me what 7-zipper line (pun intended for IT guys) meant. From there, with some ardent wrestling the logic unfolded and unraveled. Again, beautiful puzzle. Didn't watch Simon until the end, sorry Simon. Got too late. Still thanks!
Yes definitely include a few zipper lines in the app. Also please at least one lockout lines puzzle
Yes for zipper lines puzzles in the lines app / pack!
Yes! I would love zipper lines in the app!
23:00 Simon finding beautiful crazy stuff you are talented simon what an amazing puzzle
Very nice puzzle! Finished in 17:46. I saw that r4c4 had to be 7 in another way than Simon. I started as Simon by seeing that c4r123 adds to 11, so if r4c4=6 then r4c123 adds to 7. That sum is 124 and that goes with 542 (to make 6-sums) on the line in box 2. But now both 2 and 4 are on both parts of that line, and therefore 2 and 4 goes in r5c6 and r6c5. That is problem because 2+4 is not 8.
Really pleased to see Simon break in with the issue of the 1-2-4 triple in box 4 clashing with box 7. I figured the same thing out, but it took me 50 minutes to get to that point so I thought I must be missing something. It was reassuring to see Simon go through exactly the same process I did (albeit much faster).
Then I read the comments and realised I *was* missing something 😀. I didn't know about the 4x4 and 5x5 set theory. I'll have to remember that one!
@MrGrog90 Thanks for a brilliant puzzle. Really satisfying to solve, and good to have now learned a new bit of Sudoku theory too.
Thanks! The SET way is by far the the easiest route because it also allows you to show that there are no repeated digits on the 7 zipper if you think about the actual digits (and not just the sums) in the two areas. Once there are no repeats, the digits making up the 11 sum inbox 4 are fixed and you can get to 32:07 pretty quickly.
The way to show that there are no repeats is as follows: one seven on the 7 zipper => two sevens on the 8 zipper => two ones on the 8 zipper => one one on the 7 zipper => one six on the 7 zipper => two sixes on the 8 zipper => ... . You can iterate this for a bit to see that each digit from 1 to 7 must appear exactly once on the 7 zipper and exactly twice on the 8 zipper.
"And that should be a 4 by maths" Simon says while placing the very last digit in the grid 😂
36:31 What?! Does Simon really stop looking at the zipper lines to do sudoku? That’s a first I guess 😂
I'd love zipper lines in the app. They mess with my brain though. I keep trying to treat them as arrows!
At 22:22 after Simon said "if these [R4 in box 4] add up to 7, then these [C2 in box 7] add up to 7" I expected him to continue with "and that's impossible because the only way to form 7 in 3 cells is with 1/2/4, and there would be no fill for R4C2" which would prove the sum is 10 and R4C4 is 7.
He figured that out about a minute later, though it took another minute to enunciate it.
Actually I found R4C4 to be 7 using the set-theory result that the digits in the upper-left 4x4 square of cells are identical to the digits in the lower-right 5x5 square of cells, less one set of 1-9. If you just exclude box 9 and look at the total of these cells you have the 8 in R5C5, 7 pairs of cells each adding up to 8, and the 9 in R6C6, for a total of 73. Subtract 45 for box 1 and you get the zipper line centered on R4C4 sums to 28, and this value is 4 times the zipper sum which is thus 7.
67:04, had to undo a bit of work twice due to errors I made, not counting multiple mistypes. But it was dang fun using some logic I learned watching this channel, would not have been able to solve this a few years ago.
That was very interesting. Having first worked on the 9's, I then went to the set theory that box 1 and the 'ring' around it equal the two 'rings' around box 9 and it flowed beautifully from there. Finished in 21:46. Watched the start of the vid and realising Simon wasn't going to use set, decided to do it again without set.
Took 41 minutes that way as I only entered numbers when I could 'prove' the logic. Didn't flow nearly as well either. Interested to know whether gdc set this with set theory in mind...?
I really should read ALL of the comments first! 😉
I see it was intended. Great puzzle!
Excellent puzzle and solve. Solved it in 64:30. Calling it approachable is an overstatement though.
Is the intro Portobello Belle?
Fascinating to see this one solved after trying it. I did immediately think of Phistomefel, or rather the Aad's Theorem version of it, comparing the upper-left 4x4 with the lower-right 5x5 minus box 9. You get a sum of 73 in the lower-right once you place the 9, and I used that to determine the 7 in box 5, and also to argue there had to be exactly two copies of each digit from 1-7 in the lower-right set. The 7 line in the top left would allow a pair of duplicates, but only if it was a pair that summed to 7, while the lower-right L-shaped region would allow a third pair of digits summing to 8, but no pair of digits adds both to 7 and to 8. So the 16-cell regions from that set theory argument had to have exactly one 9, one 8, and two each of every digit 1-7 in order to satisfy both the set theory and the two relevant zipper lines.
The rest of the solve does seem to follow a lot of the same arguments, I just ran into them in a somewhat different order. I guess there was in fact a way around using Phistomefel that a lot of others found too, but I'm pretty pleased with having successfully used a SET argument for the first time in a solve lol, even if it took me some 66 minutes to tackle.
40:24
Very clever and lots of fun.
I got r4c4 using Aad's theorem, which this puzzle seemed to be screaming to me. Cool to see there's another way though.
I also tried using Aad's SET theorem for the break-in. I could only get r4c4 down to either 7 or 8 though. Until I belatedly saw that 9s were heavily restricted by the other zipper lines.
Really fun puzzle. Used a bit of wet theory to figure out what values to go in the centre of the zipper lines in box 5.
Thanks Simon for teaching me that "trick" haha
Is "wet theory" what you call Aad van de *Wet*ering SET theory, or is that just a coincidental typo? 😂
@@RichSmith77 haha just a typo haha
We're never in dire straits when we get a guitar intro :P
Set maybe could have been used but Aad's trick. Anyway the interactions of them clues was more like pure comedy. Absolutely loved that puzzle and im going to wait a fortnight, after that i will give the puzzle a go and see how much i can remember.
For an approachable puzzle I struggled horribly, I missed the 235 deduction in box 4 which was enough to bamboozle me for some time.
I like it when simon gives us a break from beep beep boop
"I'm doing this in the most bizarre way, I feel."
That means you're doing it right. It's a bizarre puzzle.
Have we seen a zipper+fog yet? I am sure Simon would love that.
There is one foggy zipper on the channel already. The puzzle is called "Lavender in the Fog" and the video title is "THIS Actually Makes Learning Maths FUN".
124:38, hint @myself: high numbers and how many different numbers must there be on the r4/c4 line?
That is a good looking cake!
Simon... it's really difficult for us when you say "I'll come back to that later" and then you take forever to return to it. Many times, you forget, and it takes forever. Meanwhile, it's all we can think about, and we have trouble enjoying what you're currently doing because that's eating at the back of our mind!
This seemed extremely difficult but you solved it like it was easy
44:53 ... one of the harder 'zipper' sudokus I've tackled as of late
Nice puzzle!
Wow, I'm on a streak that is hot!
These zipper lines just hit the spot!
And it's strange but it's true
i get through before you...
(sheepishly) ...'Cause I hit the pause button a lot.
(Leaves a Beach Bears Christmas album and flies off)
18:30 You could represent this by coloring the OTHER leg of that 8 in R7C4... I mean it wouldn't directly "help" per se... :D But representation it is!
My time today: 29:45. I was solver number 2293. Quite a fun puzzle!
Fun puzzle, got stuck in box 9 for a few minutes! took me 01:22:49
Zippers with doublers would be interesting as a ruleset
Took me 92:41. I wouldn't call this approachable but it was a fun solve, if very difficult. I found some of the necessary logic like pulling teeth, determining how the sets that total 9 had to be aligned in the first three columns wasn't easy for me to catch.
I learned a lot from that solve
81:56. I spent a crazy amount of time at the end pencil marking. I forgot to continue working the zipper lines.
Beautiful puzzle again! :-)
Simon, you mentioned the name Phistomefel. I may have missed something, but probably I haven't seen a Phistomefel puzzle on this channel for ages. Is he still active?
Nice zipper puzzle! Solved in 00:20:27
As always absolutely great time from you!! 🙂
@@davidrattner9 Thanks
20:45 for me. I keep struggling with zipper lines, not the biggest fan of this new rule.
Nice to see you having “normal” solving times now and then… sure you’ll get the hang of it soon, though!
Is it possible to still go over all the rules even if you previously went over them? I clicked the rules bookmark after opening the video but had to go back and try to find where you discussed the zipper lines because I thought I understood it, but I wasn't 100% sure. Usually, I click on the rules bookmark, watch them, then try to solve myself.
@ 32:33 - "I'm desperately looking" - You just pencil-marked 3s in box 1, and claimed that it could go on the line with 6 in R2C1. That's not possible because the 6 in box 4 pushes 6 into C1 in box 7. Therefore both 3 and 6 are in C2 in box 1, and neither can go on the line, so they form a 63 pair. This makes R1C2 a 45, so R2C1 is also 45. You now have a 45 pair in R1, which puts a 45 pair next to the 9 in box 3.
That was a treat of a puzzle. My method was initially the same, placing 9s and 8s, and the 7 in box 5. Then I realised the 1 had to be on the 10 part of the 7 line, so 1 had to be on the line in box 6, so 7 had to be on the line in box 8. Lots of similar deductions gradually filled out the lines, which left me with a host of triples. The few known digits could be used to gradually resolve everything.
Are the cracking the cryptic volume 1 and 2 books still available for purchase? I'm having trouble finding where to buy them
I haven't managed a zipper line puzzle. My brain insists on treating them as arrows.
I had a similar problem at first, but I found marking the center of each line with an X instead of a circle helped a bit.
Totally with you!
This is a really cool puzzle. Though I am embarrassed as to how long it took me to notice how those two lines covered that set perfectly forcing the smaller line to be a 4 + 3 sets of 8. I'm usually better at picking out that particular set, specially when it's so telegraphed like that.
Great design....manage to get all the correct sets on the lines...
But failed after 80% completed cu of sudokuing err. That ruined the solve. Mostly cuz i forgot to finish the B78 line after placing the 2nd digit that led to B9 8sum audokuing err.
how did simon miss the set where the top left 4x4 is equal to the bottom right 5x5 minus 45
Is that Nick Drake in the song intro, Simon?
Aad's theorem can save you some time :)
Zipper lines in the app please!
Got stuck bad and checked the video. The 1s at 25:21 was what I was missing... solved in 54:33 in the end.
I enjoyed the puzzle and the solve process. Please consider removing colors from a puzzle when they no longer add value. That would make it much easier for me to see. Thank you.
RUclips broke into this with ads every 8 minutes. That's too much. Now, I'm a baby boomer and I grew up with commercial TV, so it's not like I don't understand about ads. But on regular TV, the ads are prepared and predictable: they come between scenes of a drama or comedy, or in time-outs of a sports event, or between rounds of a competition or game show. On RUclips, they just barge in, interrupting Simon mid-sentence, mid-word, maybe even mid-phoneme. It's disruptive. A couple of years ago they'd at least put a little tag on screen, "ads in 5 seconds" (counting down), so you'd know it was coming. Now there's just an abrupt change of light level and sound level, with no warning.
How many ads I get seems to be fairly random; sometimes I can exit RUclips and come back later and get a better result. I don't mind ads at the start of a video, and I don't mind a block of ads in the middle of a really long one (say, over an hour), but every 8 minutes makes the videos borderline unwatchable.
Unfortunately I suspect this may be out of CTC's control. 😾
It is. In fact, it pushed me to pay for YT premium, and that paywall surely is part of the reason YT makes it so disruptive.
Yes to Zipper lines in the app
Once Simon started to paint yellow, I hoped that he will use the SET trick with squares with 4 and 5 sides
02:38:11 My first Zipper line, true challange!! Over 2 hour but managed to solve it :D Now time to watch your wideo if we did solve it same way =)
27:00 for me. That went a lot easier than yesterday's Simon puzzle!
Zipper lines don't need an odd length.
They're similar to palindromes.
An even length line would just have the same sums as they move away from the centre point.
gdc here. The way the lines are currently worded they need to be odd length. But it's true to say there is a "natural" extension of the ruleset and there is already a puzzle on LMD using it. By natural I mean, lots of people independently came up with the exact same idea of how to generalize the lines for even lengths. But I'm currently unsure if the extension adds enough to justify the increased complexity. Would you like to see puzzles using that extension?
It would make it a lot more difficult, but not necesarily more interesting. Guess we’ll know soon.
950:00 it was really fun to spend time on such a beautiful puzzle
Something’s off with your time 😂 😂 😂… 950 mn, almost 16 hours, sounds hardly possible 🤔🤔🤔
@@tessabrisac7423 nop, that's right. Its alright. (:
46:14 for me. Fantastic zipper puzzle! Kind Comment.
41:30 for me. Nice puzzle!
May I suggest that the zipper line rule be rewritten to say that “digits on a lavender ‘zipper line’ that are equidistant from that line’s center must sum to the digit in the middle of that line.” The issue with the rule as written is that it doesn’t specify that it is referencing just digits _on_ zipper lines.
Once it occurs to you that it means digits on the lines, the implication is clear, and the lines would be meaningless otherwise, but it can be misleading, and it isn’t technically correct.
40:50 and solver #3037.
I always have to laugh when you're playing a zipper line puzzle because zipper line sounds exactly like "Zipperlein" which is a German word for some slight, minor pain you're having.