Its a greatest hits showpiece with Simon displaying parity, mod 3, colouring, set theory, three in corner, apologising profusely for not seeing something incredibly complicated and then missing very simple sodoku. Bravo to both
I once had a go on a puzzle that was also five stars out of five for difficulty without any belief that I would ever solve it, but I did solve it although it took me ten hours lol.
Oh wow, this puzzle actually made it here! After I finally finished co-solving this along with Myxo and others, I was quite certain that it would have such a small chance to be here because it feels too intricate and has such a well hidden solve path but there we are, witnessing another masterwork from IcyFruit and being in awe again x) Some may wanna ask a better way for the opening so here's one that I figured out (and I think it's how the setter intended too): My idea was using SET for col 1+2 and box 7 + row 6. After some cancellation and using the fact that R5C2 = R6C3+R9C3, you're left with basically the equation 5x = 4y, and you can figure out the desired total for each of those lines. The continuation after that wasn't easy at all, but pretty much still requires the other lines' interactions along with box 2+5+8.
5x = 4y was how I solved it, too. It took me forever to even have the idea of making those two particular sets equivalent, though - Simon is so good at intuiting where and how to use SET.
@@andrewclark9809 “took forever” is also a perfect way to describe our solve too, lol. Initially, I wasn’t very keen on using SET when Myxo and companies pointed out some interesting things around box 2+5+8 with modularity and arithmetic stuff, but they didn’t really lead to a hopeful breakthrough (yet). It was spanning like hours before I switched to spotting hidden SETs and eventually got one, which then aligned perfectly with previous found results. I’m pretty certain that we would’ve stared at it for much longer time if it hadn’t been for that lucky spot.
I was about to post a comment “I feel bad for Playmaker6174” when I saw this. You solve almost every puzzle posted to LMG… and they are getting progressively more difficult and more frequent every week!
Definitely *more straightforward* than Simon's _"convolute"_ approach, but perhaps *less powerful* (see below for details). For sure, equally *awesome,* and equally *difficult* to figure out. Thanks a lot for explaining this. It was quite obvious for me that some SET trick was needed to solve, but I was not able to find it, so I used Simon's *green* and *orange* sets. The continuation was fair enough (e.g. challenging but never brutal) to make this a perfectly "balanced" puzzle in terms of gratification and difficulty.
I've solved this puzzle in a whole minute less than Simon's time. That's it, my life has peaked. I will not solve another sudoku, ever again. I'll retire undefeated with this achievement.
It's so great that after all this time, setters are still coming up with unique and interesting ideas that take even the experts over 2 hours to solve! Got some popcorn and settling in for this one ❤
"There are two states of the world, but they end up at the same position, and the point is: where is five in row six?" Could be a great intro for a sci-fi book.
I will never cease to be amused by Simon doing a 358 pencilmark, then reducing it to 58. In the same box where 8 is already present. It's hilarious to me how he can deduce so complex logic loops that would take me a year in like a minute, and then be completely oblivious to a number in the same box.
This is part of the joy of watching this like it’s a sport. You know your no athlete, you know you can’t solve it yourself but to sit and shout at a football players (or whatever sport your into) when you know deep in your soul that you couldn’t even run the length of the pitch without getting out of breath. But I can shout at Simon for missing the obvious sudoku steps at the end, when I wouldn’t have even been able to manage that break in to get to the endgame. It’s part and parcel of the joy.
If Simon's break-in was as IcyFruit intended, then I cannot fathom how this was ever set - staggeringly complexity emerging from a beautifully simple ruleset. Genius indeed.
I think Simon over-complicated the breakin. If you just take row 6 and column 3 in one set and cancel it against boxes 1 and 4 you get the relationship that 4X=5Y which leads to the 8 in box 1 and the zipper in the other set adding to 10.
Wouldn't have found that SET approach, I think, but once I saw Simon do it, I noticed he missed one step (which was marked by colouring before he removed it for doing the SET thing) that would have simplified things a bit. R6c7, the zipper complement of r6c4, has to sit in either r4c4 or r5c4, Simon had worked that out before. So if we shift this orange cell over into column 4 in one of these places (it doesn't matter into which one), we have an almost complete column in orange being equal in sum to the green set of digits, which is divisible by 5. Hence the differnece, which is the single missing digit in the column is a 5 (the column adds up to 45, which is divisible by 5, orange is equal in sum to green which is divisible by 5, so the missing digit must be as well, and there is no other sudoku digit that is divisible by 5). It follows that the orange and the green set add up to 40 each, i.e. the zipper line in boxes 1 and 2 being the green set adds up to 40, so its central digit is 8.
yes, I got quickly that in C4R4/5 must be a 1 or 5 and together with his insight that the orange in R7C6 goes also in C4R4/5 he would have seen the multiple of 5 right away...
I was delighted to see a very long video on the channel, and sorry that I could not, I knew, watch it all at once. But it was definitely a worthwhile venture to come back to it over two sessions. Thanks so much, Simon, to you and Mark for consistently bringing us the very best the sudoku world has to offer. No, I don't think I'll have a go, but I am so happy that you did and you brought me along for the ride.
Alas, if only Simon had recalled the previous (purple) coloring of r6c7 at 44:54. We could have skipped ahead to 1:05:55. But holy wow, Simon! That was a great bit of set theory to get to 44:54! I was very impressed
I couldn’t watch the video straight away, so I gave it a shot and stared at the grid for 45 minutes only making the simplest of deductions. That SET is wild and I never saw the possibilities until Simon pointed them out. Amazing puzzle Icyfruit!
At around 57:00, it's simpler to remember that the orange cell r6c7 is one of r4 or r5 in c4, so eight cells in c4 sum to the same as the green zipper in the top left, which has to be a multiple of 5. Therefore X for the green zipper is 8, and 5 is the other digit missing from orange in r4 or r5 in c4.
Around 1:31:00, you can prove the sum for the G-shaped line by asking where the 7 goes in column 3. It's not in box 1, not in row 9, and if it was in row 7 or 8, it would be opposite a second 3 in row 6. This puts it in box 4, so the 7 in box 4 can't be the center of the line. I think that may simplify some of your next steps.
that first digit is certainly one for the ages, what an absolutely insane puzzle. bravo simon for finishing this one, that was certainly brutally difficult
I like these uncomplicated puzzles, nice simple rule set. Nothing new or super weird. Just me staring for an hour, going "How? Where are you supposed to look?" "Simon? Help?" .... Oh, of course. Why didn't I think to compare 1 row, 2 columns and 3 boxes over 4 zippers so I could torture myself for the rest of the evening. Not especially easy even after that insane break-in. Well set Icy Fruit.
I've tried this puzzle for about an hour and a half, coudn't get it started no matter what I did, and the seeing the solve, now i know why. This break-in was impossible for me to do it Good job, Simon. I love your solves and yours and Mark channel is amazing! Love you guys
i'm pretty happy that i got some of the early deductions, but this used techniques that I just had never thought of. Loved every minute of this entire video.
Simon is always so conscious and worried about his hmms and pregnant pauses making the solves not enjoyable to watch. I just think they make these relaxing and rewarding to watch, like an empty space accentuating the following breakthrough.
1:32:53 finish. This was very tricky to solve, though at a certain point everything just clicked into place. Incredible puzzle! Simon, when your solve runs past a certain length, you tend to second guess a lot of logic, extending your time even further. Have faith in yourself.
At 1:25:50, that cannot be 4-7-3 because 3 & 7 have to both be on the zipper. They'd both need to be A or B, making the A,B pair simultaneously 35 and 17.
12:54 This conclusion is correct, but there is much a simpler approach yielding a much stronger conclusion. If you notice the placement of the line, you can see that the sum X must be formed by 4 distinct pairs of digits. There are only three numbers that have 4 different ways of being formed as a sum by exactly 2 sudoku digits, and those numbers are 9,10, and 11. Therefore, for this particular zipper line, you can conclude right away X is 9, 10, or 11. This way, you also know that 8 of the 9 sudoku digits must appear exactly once each within those 8 squares of the zipper line in those 3 rows, and the only digits which can be excluded from those 8 digits are the digits 1, 5, and 9.
Incredible puzzle. I needed a couple of hints from Simon's solve. 1) To consider the sets r6+c3 against box 4 and box 5. 2) Why r1c4 couldn't be 6. Both times, I could work the deductions out, but as is often the case, I need Simon's insight to know the right question to ask. Amazing solve by Simon.
It was an extremely admirable solve but I think I managed to find a really neat way through the first part (after the hint to look for SET). Once you’d used the information after cancelling to determine the 8, you could restore the original equivalence (produced by comparing boxes 1 & 4 against column 3 and row 6) and use the fact the digits were the same to go much further. Since double 4 could not go on the 8 zipper line or 4 in the centre of the zipper line with 5 on it, one of the 10 combinations was removed (4+6) and double 5 therefore had to appear as a combination for there to be four different pairs. The repeat for the zipper line in the other set was then 3,5 and the mystery zipper centre was a 9 (since the 6 for the 8 sum had to go somewhere in the one set, along with the extra 3). Very clever setting. Unfortunately, I got slowed right down by missing row 1 column 4 having to be an 8. I’m relieved that I got through without it but gutted that I didn’t see it, I even had the 2,4 pair but would honestly never have noticed the implication. I also forgot that I’d originally worked out the line at the top of column 3 was modulo 3, which was mildly annoying.
Sorry, think I messed up the order of the logic a bit there (initially at least, there could have been 3 pairs adding to 10). The digits in set A were 1,2,3,5,6,7,8 and some others. 6 therefore had to appear in set B and since 4 couldn’t, it had to go in the zipper line sum with either 1, 2 or 3. Thus, 5 had to be in a 10 sum, forcing there to be 4 different pairs and making the repeats 3 and 5.
48:55 A more elegant way to eliminate 5 from r8c8 is to remember that r6c7 must be in r4/5c4 (14:39) and to notice that orange is now an almost complete column. Green is 5x, sum of green equals sum of orange (45:20), and orange is one digit less than an entire set of 1-9 (column 4 with r6c7 transposed into one of r4/5c4), therefor the missing digit must be 5 in r4/5c4. This rules 5 out of r7-9c4 which r8c8 must be in (27:05). Therefor, r8c8 cannot be 5
I just found some alternate logic for the deduction starting around 40:00 and ending around 57:00, with the recap starting at 53:00. It relies on a bit of algebra. We can express the orange sum as S = 45 - x + z, where x = r5c4 + r6c4, and z = r5c7. We can also express x as x = 45 - y, where y is the sum of everything else in the middle box, which is conveniently all on the same "stone henge line." Then the equation becomes S = 45 - (45 - y) + z = y + z. And everything on y+z is on the line! It is exactly 4 pairs of numbers. Let's denote the sum of any pair of numbers on the "stone henge line" as n. As Simon deduced earlier, n could only be 9, 10, or 11. This means that S = 4 * n. Meanwhile the sum of orange section is also the sum of the line on the top left, which is a multiple of 5. This means that if n = 9, S = 36, doesn't work. If n = 11, S = 44, doesn't work. Therefore n = 10, S = 40, and r3c2 = 8! A similar method would be to subtract from the orange area all of column three and adding back all of the middle box, leaving the same sum. From the colored area you could find that the sum of the orange area is 4 times the sum of pairs in the "stone henge line," and the rest is similar.
Puzzle looked too complex for me but a cracking puzzle. Something that I picked up on during the video was at 1:06:00 when you put the 245 into R1C3 and R2C3, that the 1/7 has to be the sum adding to 8 that would have been A & B as put 1 or 7 anywhere else on the line would have meant the other couldn't appear in the box.
I finished in 207 minutes. This has been an absolute blast of a puzzle. The way the geometry worked between each line was tremendous to spot. I had some fear when I saw the ruleset versus the time and knew I was in for a tough one. However, everything felt very nice, even if it took me a long time to see it. I think my favorite part was realizing that the even line in box 5 had to be a 10 total, due to the remainder number being forced into r6c3, breaking the line that it is on by having two of the same number in box 7. Thus, the only way to make it work is by having a repeated digit on the even line from box 5 in row 6 and column 3. There were many more cool tricks just like that, which were also amazing to see. This has to be one of my favorites. Great Puzzle! Edit: After having watched the video, I am surprised at how different Simon's solve was from mine. I completely ignored SET Theory. I had a suspicion that it could be used, but I'm usually so bad it, that I tend to ignore it unless absolutely necessary. It turns out that it isn't necessary. The geometry between the lines are so powerful that it naturally doesn't require SET theory. I started by spotting that 5's had to belong on the even line in box 5. That led to 5's in r6c89 and r78c3. This forced dual 5's onto the odd line in box 1. This helped me so much, because it forced extra digits that had to belong in column 3 of box 7. In addition, I had noticed whatever digit accompanied 5 in column 3 of box 7 had to be included in row 6 of box 4, along with a pair that added to 10. All that together led to some tight geometry that allowed me to avoid SET Theory and reduce digits and finally place an 8 into r3c2 and a 9 into r5c2. It is quite amazing that Simon and I had such different styles, each spotting various parts way before the other. What fun!
theres a way to break in without going through cases. it uses the similar sets as simon, except without the orange col 4 but adds in instead that r6c7 is in one of r4c4 or r5c4. we can then virtually cancel them out of both sets. this means there's nothing left in orange, and the top left zipper plus one of r4c4 or r5c4 left in green. since green had one extra set of the digits 1-9, they add to the secret. also all of the greens on the zipper line add to a multiple of 5, and the only one that leave the last green cell as a digit from 1-9 is 8, with the last green = 5.
I could never have spotted the set logic on my own, but after watching this video I did see a slightly different way to look at it. Highlight boxes 1 and 4 in one color. Then highlight column 3 and row 6 in another. Do the overlapping cancelations, remember that R6C3 is double counted. Then cancel out R5C2 with R6C3 and R9C3. What you end up with is the same set Simon had in green (call it 5y) vs four sets of the other color (call it 4x). So now you know you need a total that is divisible by 4 and 5. In my mind it is a little less convoluted than the set Simon ended up with.
This took me three months to solve. I opened the puzzle in a tab the day this video was published, then I kept looking at it occasionally without figuring out how to break in. I thought about sets, but I just couldn't come up with anything that worked, and especially not something as convoluted as what Simon did here. I ended up focusing on the zipper line that occupies most of box 5 and was able to rule out 11 because that would force the two outer cells on both ends into the upper left zipper and the middle cell of the one below it. Then I checked every possible combination of digits they could be and nothing worked, because it usually resulted in a repeated 9 in row 5. The same logic also ruled out 10 as the sum with no 5 in the outer cells. I got the 8 in box 1 by figuring out two digits would have to be absent from that zipper, and these digits would have to appear in the three cells below the zipper in box 4, and the only combination of digits that worked was 9 and 4. I had already figured out the upper left zipper would be an 8 or 9, and there was no combination of digits summing to 9 that would be able to repeat below it.
so proud to have done this in 86:25! and the break-ins are really thought provoking: it's the importance of numbers 159, what do you put in the cell r6c3, where does 4 go in column 4.
very nice puzzle for such a simple ruleset. One can deduce from the start that the sum of the line in the center box must be bigger than 9 since 9 in box 5 must be on it.
I wonder if that set logic followed by Simon was the intended "break in" or IcyFruit had some simpler solve path in mind Either way, finding any path through the puzzle is quite an achievement
Simon always notices and deduces a bajillion times more than I do, so it's particularly thrilling when I see something he doesn't. "Where is 5 in column 3?" leads to be bunch of stuff. He did eventually get to the very similar, "Where is 5 in row 6?" question.
You could have saved 20 minutes at the start just by asking where 9 goes in box 5. It can't go on the short segment, because that line sums to a single digit, so the long segment has a sum of at least 10. It can't be higher than 11, because it has to have four ways to make it, with no doubled digits. In box 9, that line also needs four ways, so it sums to 9, 10, or 11, with the centre being 9, 5, or 1 respectively. That digit must appear in R7/9C4, and therefore must be on the long segment in box 5. This means the line sums are different. R6C7 must appear in R4/5C4 along with the digit which isn't used to make the four pairs, which must be 1 or 5. I didn't use SET per se, but my head calculations had the same effect as your SET. That allowed me to work out that the long segment in box 5 must have a sum of 10. This is because the three sets of three cells in C4 are some permutation of 12, 15, and 18. The top + bottom + X must be divisible by 5. The only combination which works is 12 + 18 + 10. @ 1:27:22 - "How am I going to do this?" - Ask where 5 goes in C3. If it's in box 4, it aligns with the 5s in box 5, so pushes 5 onto the 10-line in box 6, where it also needs a 5 on the other end, clashing with the hypothetical 5. Therefore there is a 5 in R7/8C3 and R6C8/9. This forces 5 onto the hook in box 4, which forces 3 into the 3 cells at the other end, and therefore not in R2C2/C3C1, so they are a 17 pair. @ 1:44:21 - You know 356 are on the line in box 1, so you know 235 are on it in box 4, ruling 36 off the diagonal. This places 7 in R5C3, 6 in R6C3, and 3 in R9C3 @ 1:49:15 - You're worried that you can't recall why you pencil-marked 2s into C1 in box 1 - The whole point of pencil-marking is that once you've made a deduction, you write it down and can forget about how you worked it out, freeing your mind to think about something else. If you're going to mistrust them, there's not really much point in making them. You might as well just work it out from scratch repeatedly. @ 1:59:48 - "And now if that's a 5" - how can it be? It's part of a 58 pair where the other is in a box with an 8. Don't make rubbish pencil-marks. You work on some logic, and put in pencil-marks without checking whether other rules (like sudoku) resolve them. You waste a lot of time considering stuff that could have been ruled out when you made the pencil-marks. @ 2:06:40 - "It is 6, 7, or 9" - You've just done it again, making rubbish pencil-marks. There's a 46 pair immediately to the left, it can't be 6. @ 2:07:37 - "that's 8, or 8 is there" - And again! There's an 8 above your second pencil-mark. You could resolve both the 8 and the 3, if only you looked. This was quite tough in places, but very interesting. I never got stuck, but there were spells where a sequence of very minor deductions just kept things moving on. Suddenly, it achieved critical mass and gave in.
When you spot something before Simon does and he then spends the next 10 min going over incredibly complex stuff and you're like "you were so close to seeing it!" And then he finds something else that I never would have spotted on my own. lol
I think it is easier to see this puzzle with odd/even polarity. The larger lower line is mod 3 AND mod 4. R1C4 then becomes very constrained, and has a parity affect on C4. Also look at how R5C2 interacts with purple. Near the end, look at how many evens are available in box 2, and what that implies for the length 6 line.
In 15 seconds I got to where Simon got to in 15 minutes, then in the next 2 hrs Simon got to where I will never get to. I'm glad I gave up on this one. I don't mind spending a week on a puzzle. I've done that many times with puzzles on LMG, but I would never have figured this one out.
There's a (slightly) easier way to get the first digit. Once you get the SET coloring pattern equating the top left zipper line to most of column 4 + r6c7, You can use your earlier observation that the central zipper line forces the digit from r6c7 into either r4c4 or r5c4, which are the two cells from column 4 not already in the orange set. That means that the green set sums to 45 minus one digit. 40 is the only possible multiple of 5 satisfying that condition, giving the 8 in r3c2.
Wow. Well, I had to come back to the video a couple times to get a prod; I was looking for the set theory quite a bit early on, and I was in the right general place, but I didn't find the sets --- quite the insight there. And I got stuck later winnowing down r1c4 because I couldn't see the conflict, which was just a matter of not being comfortable with the logic with these lines, I think. A very educational puzzle even in "defeat". 170:49
Whenever I see these videos published, I normally pause the video and then attempt the solve, and either watch the video after I solve it, or let is go on in the background if I get stuck. This puzzle I started the solve, and was immediately attracted to the lines in box-9. I then somewhat quickly determined that the cell in the middle of that box had to be 1, 5 or 9... then started looking for the next path.. but couldn't see it immediately... so I then started looking elsewhere... mostly box-1... and started looking at the options there. Wasn't seeing an easy path there, so I then glanced at the video to see how long the video was... 2 hours?!!! oh boy... if it takes him 2 hours to solve it... will probably take me 10 hours (if I even can...)... so I then decided to just watch the video of Simon solving it (haven't watched it all yet... at 16:32 so far, be interesting to see what amazing tricks needed to solve this).
To add a little more clarity to my statement about "the next path". once I noticed the single cell in box-9 that wasn't on a line, and then noticed there was 2 other boxes that also only had one cell with no line in it (plus box 4 which also caught my attention)... that is what I was looking at guessing that it was the path to follow... but couldn't see anything immediate.
Yep. I spent most of the first 10 minutes of the solve telling the screen "9 has to appear in this box somewhere which means X is at least 10, and r6c7 appears in r4/5c4 so X can't be bigger than 11 or the box total will be over 45!"
Simon, having watched your geometry analysis I realise there is a far easier way by considering boxes 1 and 4 with column 3 and row 6. Then after canceling the centre digit of box 4 to leave just the “ multiple of 5” zipper as you did, the other set is simply 4 pairs so since the total must exceed 20, the only other multiple of 4x5 is 40 hence the 8 in box 2 and the 10-sum dominoes for box 5. Of course it’s easy to see once you’d suggested the “set” method
You started trying lines vs boxes but then had to try a lot of cases before reaching a conclusion. However, if you had just added the rest of a box to both sets the answer crystalizes simply. Well, you'd have to know where the furthest right cell is in the other set and the possible values of the final cell, but luckily that I had before I started looking at this. I hope that's clear enough without being too clear.
Had this thought a while ago, but since you mentioned a 600k subscriber celebration, I think the best idea for a special video would be to solve a sudoku while flying with Maverick. Maybe save it for 1mil!
Gotta finish watching this today since I fell asleep last night after only an hour and 10 minutes ( I think Simon got one digit) All I remember is that the madman on the screen started shouting about set theory nobody's ever seen before. Then I dreamed I went to Surrey for a wedding and we were all waiting on Simon to finish his puzzle before we could start the festivities, but nobody knew when the puzzle would be solved. 😅 Strange things in the world of fever dreams and logic puzzles indeed.
I got pretty quick to the point that the line in box 5 wasn’t a 9 line, as the line in box 4 couldn’t exceed 9, meaning that the box 5 line must have 9 on it (otherwise you would have a 9/0 pair on the box 4 line). Then it was a screaming halt from there…
I think it would have helped Simon a lot to get rid of the colouring once the colouring served its purpose. This would also make it easier to recycle colors elsewhere. It would also help make the grid look nicer, since a lot of the times, the colouring can make the grid look very jarring.
Simon apologizing for the length of a video with an epic solve that most of the audience will have enjoyed immensely is the most Simon thing ever. That and totally overlooking obvious sudoku eliminations making us yell at our screens. 😜
I'm very happy to see this one done by Simon, as I tried it... for hours. And failed it. I found most of the first elements, the 9 10 11 sums, putting 1 5 or 9 in column 4 and preventing the sums to be the same. I found the modulo 3 thing in line 2. But, my god, that SET. I was wandering if there are other ways to disambiguate it all. Was that set really mandatory to get it through ?
Love these monstrous puzzles. Nearly six and a half hours for me. But for someone who doesn't like pencil marks cluttering up the grid, there is an awful lot of superfluous colouring in Simon's grid.
I love the fact, that you can not have a 6 on the "bulb" in C4 because you would have a double 2 on the line but can only have an 8 on the "bulb" if you have a double 4.😁
wow I kind of tuned out around 30 minutes. Sometimes I don't mind long puzzles, but this isn't one of them. I came back at 1 hour 30 and......you have 3 digits. lol that's crazy. I basically tuned out because I was completely lost.
I will say that the break-in would be cleaner if you moved the final set of orange cells in column-4 into box-5. Then you are left with 5*green=4*orange.
an easier way of seeing how the zipper in box 5 had to add up to 10 is to note that it needs 5 ways of adding up, as r6c3 will end up seeing all 4 ways of adding to 9 and 11, therefore repeating on its own zipper line in box 5, and you end up with a repeated digit on the zipper line in box 7
Edit: Ignore what I originally wrote. Someone else has written a similar comment, and it has finally sunk in that the cell you are talking about (r6c3) is not r6c4. r6c3 does indeed see four ways of making X. Apologies.
@GiantPeccan replied in another comment that this isn't correct. The pairs summing to X formed by (r6c8, r7c3) and (r6c9, r8c3) could be the same pair, just reversed. So r6c3 doesn't have to be seeing four different ways of making X along row 6 and down column 3.
About 2 hours for me, not sure because I left the timer running for a while. But I cheated a bit. I had decided not to try it and then saw a comment about the set break in giving a 4x=5y equation. I opened the puzzle to see it and the ended up solving it. It was really fun
Im still racking my brain to figure out why C3R1-2 needs to be %3. Wouldnt something like 1-4-7-3-6 have filled out the line at the time of that deduction? I feel like the 2-4 pair there got stumbled on. I would have never been able to come close to solving these, kudos to featuring it. Im just feeling like im missing the logic
I didn't get very far before I decided to just watch SImon do his wizardry, but it did seem to be pretty clear to me from the get-go that the centers of the two lines in boxes 1 and 4 had to be 8 and 9, since by the geometry of the two zipper lines, there needed to be 4 distinct ways of adding pairs up to the center value. Was I making it too easy?
He ends up only using three ways to make the 8 on the box 1 zipper though, since 4+4 is ruled out. And if it's possible in 3 ways, I don't think there's an easy way to rule out 7.
One thing I appreciate is that you in the end apologize to people who want to have a more manageable schedule and approachable puzzles instead of monstrous tasks every night.
Houston, we have a problem... Although I do genuinely like the extensive movie-length solve videos very much, I have to confess that it becomes increasingly difficult for me to watch them in practice. I could often afford to do so during lockdown times, but far less so in "normal times", given other responsibilities. If there are too many long videos in one week, I often take my refuge to re-watching older CtC videos, but most of them I know already. ;-) So I am glad lately whenever I spot relatively short new videos. Maybe I'll try to watch the present solve bit by bit in the upcoming week, for it looks quite interesting all the same. Meanwhile, I keep being blown away by all the effort, love and spirit being invested in all these solves by Mark and Simon day after day!
speaking of the app.......I love the latest update where you can apply empty stars. I just wish that feature was on ALL the apps. Or better yet.....Please re-release all the old puzzles as DLC on the new app. I don't mind buying them again on both android and steam. That's fine. I just hate that I don't have the empty stars. Also android has many updates for the old apps (not including empty stars) using the new UI for the CTC app. But steam didn't get those updates. This is why I would love to just have everything on a single app.
Column 1,2,3 and row 6. Box 1,4,5,7. Set. Yes... That makes row 6 column 789 equal 6 cells in box 5. 3 cells sum to at least 23 . So you can polarity shade.. hang on .. 😮 think im wrong. Sorry 😟
Its a greatest hits showpiece with Simon displaying parity, mod 3, colouring, set theory, three in corner, apologising profusely for not seeing something incredibly complicated and then missing very simple sodoku. Bravo to both
Simon: Do have a go on the puzzle. I see it takes Simon over 2 hours. Me: No I don't think I will.
Yeah Simon, best I can do is having a watch of YOU having a go. It'll be more fun that way for all of us lol.
I once had a go on a puzzle that was also five stars out of five for difficulty without any belief that I would ever solve it, but I did solve it although it took me ten hours lol.
I heard that in the old-man Steve Rogers' voice, which is fitting, as that's how old I'd probably be to solve a Simon-took-2-plus-hours puzzle.
That's exactly why I will. Those puzzles are the best and give amazing satisfaction when you manage to beat them. Might take a week for me, though.
My rule for if I do a puzzle or not is if the video is under an hour then I'll attempt it. 😅
Friend: What'd you watch last night?
Me: A brutal two hour long fight.
Friend: Oh, a movie like Raging Bull?
Me: No. Longer. A sudoku.
Oh wow, this puzzle actually made it here! After I finally finished co-solving this along with Myxo and others, I was quite certain that it would have such a small chance to be here because it feels too intricate and has such a well hidden solve path but there we are, witnessing another masterwork from IcyFruit and being in awe again x)
Some may wanna ask a better way for the opening so here's one that I figured out (and I think it's how the setter intended too):
My idea was using SET for col 1+2 and box 7 + row 6. After some cancellation and using the fact that R5C2 = R6C3+R9C3, you're left with basically the equation 5x = 4y, and you can figure out the desired total for each of those lines.
The continuation after that wasn't easy at all, but pretty much still requires the other lines' interactions along with box 2+5+8.
Simon seems to have found a less arduous way through than I did, but I am in awe of what you found. That needs to be seen too ...
5x = 4y was how I solved it, too. It took me forever to even have the idea of making those two particular sets equivalent, though - Simon is so good at intuiting where and how to use SET.
@@andrewclark9809 “took forever” is also a perfect way to describe our solve too, lol. Initially, I wasn’t very keen on using SET when Myxo and companies pointed out some interesting things around box 2+5+8 with modularity and arithmetic stuff, but they didn’t really lead to a hopeful breakthrough (yet). It was spanning like hours before I switched to spotting hidden SETs and eventually got one, which then aligned perfectly with previous found results. I’m pretty certain that we would’ve stared at it for much longer time if it hadn’t been for that lucky spot.
I was about to post a comment “I feel bad for Playmaker6174” when I saw this.
You solve almost every puzzle posted to LMG… and they are getting progressively more difficult and more frequent every week!
Definitely *more straightforward* than Simon's _"convolute"_ approach, but perhaps *less powerful* (see below for details). For sure, equally *awesome,* and equally *difficult* to figure out. Thanks a lot for explaining this.
It was quite obvious for me that some SET trick was needed to solve, but I was not able to find it, so I used Simon's *green* and *orange* sets. The continuation was fair enough (e.g. challenging but never brutal) to make this a perfectly "balanced" puzzle in terms of gratification and difficulty.
I've solved this puzzle in a whole minute less than Simon's time. That's it, my life has peaked. I will not solve another sudoku, ever again. I'll retire undefeated with this achievement.
It's so great that after all this time, setters are still coming up with unique and interesting ideas that take even the experts over 2 hours to solve! Got some popcorn and settling in for this one ❤
I wish I had popcorn instead of dishes to do - but this great long serving of Simon's masterful mind will certainly make it feel much less like work!
"There are two states of the world, but they end up at the same position, and the point is: where is five in row six?" Could be a great intro for a sci-fi book.
Simon: Do have a go on the puzzle.
Me finding the video is 2 hours long: Well, I guess I will find popcorn and cola, and just enjoy Simon's solve.
Or in my case: Nah, I have to sleep tonight so maybe some other time.
I remember when a 40 minute video was considered a marathon solve.
@@57thorns I miss the day that a 1 hour and 28 minutes video was defined as a movie. Look how far we reach from there.
I will never cease to be amused by Simon doing a 358 pencilmark, then reducing it to 58. In the same box where 8 is already present. It's hilarious to me how he can deduce so complex logic loops that would take me a year in like a minute, and then be completely oblivious to a number in the same box.
It's tunnel vision
This is part of the joy of watching this like it’s a sport.
You know your no athlete, you know you can’t solve it yourself but to sit and shout at a football players (or whatever sport your into) when you know deep in your soul that you couldn’t even run the length of the pitch without getting out of breath.
But I can shout at Simon for missing the obvious sudoku steps at the end, when I wouldn’t have even been able to manage that break in to get to the endgame. It’s part and parcel of the joy.
The problem with seeing dimensions that we can't see is that his view of our dimension is cloudy for him
If Simon's break-in was as IcyFruit intended, then I cannot fathom how this was ever set - staggeringly complexity emerging from a beautifully simple ruleset. Genius indeed.
I think Simon over-complicated the breakin. If you just take row 6 and column 3 in one set and cancel it against boxes 1 and 4 you get the relationship that 4X=5Y which leads to the 8 in box 1 and the zipper in the other set adding to 10.
Wouldn't have found that SET approach, I think, but once I saw Simon do it, I noticed he missed one step (which was marked by colouring before he removed it for doing the SET thing) that would have simplified things a bit. R6c7, the zipper complement of r6c4, has to sit in either r4c4 or r5c4, Simon had worked that out before. So if we shift this orange cell over into column 4 in one of these places (it doesn't matter into which one), we have an almost complete column in orange being equal in sum to the green set of digits, which is divisible by 5. Hence the differnece, which is the single missing digit in the column is a 5 (the column adds up to 45, which is divisible by 5, orange is equal in sum to green which is divisible by 5, so the missing digit must be as well, and there is no other sudoku digit that is divisible by 5). It follows that the orange and the green set add up to 40 each, i.e. the zipper line in boxes 1 and 2 being the green set adds up to 40, so its central digit is 8.
yes, I got quickly that in C4R4/5 must be a 1 or 5 and together with his insight that the orange in R7C6 goes also in C4R4/5 he would have seen the multiple of 5 right away...
Just the fact that the first placed digit is in r3c2 and it is placed looking at r8c8 shows how clever and intricate this puzzle is. Astonishing
I was delighted to see a very long video on the channel, and sorry that I could not, I knew, watch it all at once. But it was definitely a worthwhile venture to come back to it over two sessions. Thanks so much, Simon, to you and Mark for consistently bringing us the very best the sudoku world has to offer. No, I don't think I'll have a go, but I am so happy that you did and you brought me along for the ride.
Stunning puzzle, looking forward to the solve!
This is one sudoku of all time
it sure is
Alas, if only Simon had recalled the previous (purple) coloring of r6c7 at 44:54. We could have skipped ahead to 1:05:55. But holy wow, Simon! That was a great bit of set theory to get to 44:54! I was very impressed
Brilliant puzzle. It took me longer than Simon, i think, but delighted just to have finished it. Thoroughly enjoyable.
I couldn’t watch the video straight away, so I gave it a shot and stared at the grid for 45 minutes only making the simplest of deductions. That SET is wild and I never saw the possibilities until Simon pointed them out. Amazing puzzle Icyfruit!
At around 57:00, it's simpler to remember that the orange cell r6c7 is one of r4 or r5 in c4, so eight cells in c4 sum to the same as the green zipper in the top left, which has to be a multiple of 5. Therefore X for the green zipper is 8, and 5 is the other digit missing from orange in r4 or r5 in c4.
Around 1:31:00, you can prove the sum for the G-shaped line by asking where the 7 goes in column 3. It's not in box 1, not in row 9, and if it was in row 7 or 8, it would be opposite a second 3 in row 6. This puts it in box 4, so the 7 in box 4 can't be the center of the line. I think that may simplify some of your next steps.
Very clever (once I'd worked which was the G-shaped line 🙂)
that first digit is certainly one for the ages, what an absolutely insane puzzle. bravo simon for finishing this one, that was certainly brutally difficult
Congratulations on 575,000 subscribers! It should be at least 100x that! Share this with your friends, folks!
I like these uncomplicated puzzles, nice simple rule set. Nothing new or super weird.
Just me staring for an hour, going "How? Where are you supposed to look?"
"Simon? Help?" .... Oh, of course. Why didn't I think to compare 1 row, 2 columns and 3 boxes over 4 zippers so I could torture myself for the rest of the evening.
Not especially easy even after that insane break-in. Well set Icy Fruit.
LOL, It's simple really, once you get down to it :P
I've tried this puzzle for about an hour and a half, coudn't get it started no matter what I did, and the seeing the solve, now i know why. This break-in was impossible for me to do it
Good job, Simon. I love your solves and yours and Mark channel is amazing! Love you guys
Just astounding! The setting was otherworldly and the solve was superhuman!!!
i'm pretty happy that i got some of the early deductions, but this used techniques that I just had never thought of. Loved every minute of this entire video.
That is just a beast of a puzzle! Great work IcyFruit and well done Simon for finishing it!
Simon is always so conscious and worried about his hmms and pregnant pauses making the solves not enjoyable to watch. I just think they make these relaxing and rewarding to watch, like an empty space accentuating the following breakthrough.
1:32:53 finish. This was very tricky to solve, though at a certain point everything just clicked into place. Incredible puzzle!
Simon, when your solve runs past a certain length, you tend to second guess a lot of logic, extending your time even further. Have faith in yourself.
At 1:25:50, that cannot be 4-7-3 because 3 & 7 have to both be on the zipper. They'd both need to be A or B, making the A,B pair simultaneously 35 and 17.
12:54 This conclusion is correct, but there is much a simpler approach yielding a much stronger conclusion. If you notice the placement of the line, you can see that the sum X must be formed by 4 distinct pairs of digits. There are only three numbers that have 4 different ways of being formed as a sum by exactly 2 sudoku digits, and those numbers are 9,10, and 11. Therefore, for this particular zipper line, you can conclude right away X is 9, 10, or 11. This way, you also know that 8 of the 9 sudoku digits must appear exactly once each within those 8 squares of the zipper line in those 3 rows, and the only digits which can be excluded from those 8 digits are the digits 1, 5, and 9.
Incredible puzzle.
I needed a couple of hints from Simon's solve.
1) To consider the sets r6+c3 against box 4 and box 5.
2) Why r1c4 couldn't be 6.
Both times, I could work the deductions out, but as is often the case, I need Simon's insight to know the right question to ask.
Amazing solve by Simon.
It was an extremely admirable solve but I think I managed to find a really neat way through the first part (after the hint to look for SET). Once you’d used the information after cancelling to determine the 8, you could restore the original equivalence (produced by comparing boxes 1 & 4 against column 3 and row 6) and use the fact the digits were the same to go much further. Since double 4 could not go on the 8 zipper line or 4 in the centre of the zipper line with 5 on it, one of the 10 combinations was removed (4+6) and double 5 therefore had to appear as a combination for there to be four different pairs. The repeat for the zipper line in the other set was then 3,5 and the mystery zipper centre was a 9 (since the 6 for the 8 sum had to go somewhere in the one set, along with the extra 3). Very clever setting. Unfortunately, I got slowed right down by missing row 1 column 4 having to be an 8. I’m relieved that I got through without it but gutted that I didn’t see it, I even had the 2,4 pair but would honestly never have noticed the implication. I also forgot that I’d originally worked out the line at the top of column 3 was modulo 3, which was mildly annoying.
Sorry, think I messed up the order of the logic a bit there (initially at least, there could have been 3 pairs adding to 10). The digits in set A were 1,2,3,5,6,7,8 and some others. 6 therefore had to appear in set B and since 4 couldn’t, it had to go in the zipper line sum with either 1, 2 or 3. Thus, 5 had to be in a 10 sum, forcing there to be 4 different pairs and making the repeats 3 and 5.
Ok, I tried solving again and was amazed to find it took me just over 30 minutes. This puzzle was definitely a lesson in efficiency.
Chapeau l'artiste. Seeing your thoughts going on is just mindblowing!
48:55 A more elegant way to eliminate 5 from r8c8 is to remember that r6c7 must be in r4/5c4 (14:39) and to notice that orange is now an almost complete column. Green is 5x, sum of green equals sum of orange (45:20), and orange is one digit less than an entire set of 1-9 (column 4 with r6c7 transposed into one of r4/5c4), therefor the missing digit must be 5 in r4/5c4. This rules 5 out of r7-9c4 which r8c8 must be in (27:05). Therefor, r8c8 cannot be 5
I want a video of Simon reacting to this video some day to be in the audience's shoes.
I just found some alternate logic for the deduction starting around 40:00 and ending around 57:00, with the recap starting at 53:00. It relies on a bit of algebra.
We can express the orange sum as S = 45 - x + z, where x = r5c4 + r6c4, and z = r5c7.
We can also express x as x = 45 - y, where y is the sum of everything else in the middle box, which is conveniently all on the same "stone henge line."
Then the equation becomes S = 45 - (45 - y) + z = y + z.
And everything on y+z is on the line! It is exactly 4 pairs of numbers. Let's denote the sum of any pair of numbers on the "stone henge line" as n. As Simon deduced earlier, n could only be 9, 10, or 11.
This means that S = 4 * n.
Meanwhile the sum of orange section is also the sum of the line on the top left, which is a multiple of 5.
This means that if n = 9, S = 36, doesn't work. If n = 11, S = 44, doesn't work. Therefore n = 10, S = 40, and r3c2 = 8!
A similar method would be to subtract from the orange area all of column three and adding back all of the middle box, leaving the same sum. From the colored area you could find that the sum of the orange area is 4 times the sum of pairs in the "stone henge line," and the rest is similar.
I'm holding Icy Fruit's beer.
Puzzle looked too complex for me but a cracking puzzle. Something that I picked up on during the video was at 1:06:00 when you put the 245 into R1C3 and R2C3, that the 1/7 has to be the sum adding to 8 that would have been A & B as put 1 or 7 anywhere else on the line would have meant the other couldn't appear in the box.
I finished in 207 minutes. This has been an absolute blast of a puzzle. The way the geometry worked between each line was tremendous to spot. I had some fear when I saw the ruleset versus the time and knew I was in for a tough one. However, everything felt very nice, even if it took me a long time to see it. I think my favorite part was realizing that the even line in box 5 had to be a 10 total, due to the remainder number being forced into r6c3, breaking the line that it is on by having two of the same number in box 7. Thus, the only way to make it work is by having a repeated digit on the even line from box 5 in row 6 and column 3. There were many more cool tricks just like that, which were also amazing to see. This has to be one of my favorites. Great Puzzle!
Edit: After having watched the video, I am surprised at how different Simon's solve was from mine. I completely ignored SET Theory. I had a suspicion that it could be used, but I'm usually so bad it, that I tend to ignore it unless absolutely necessary. It turns out that it isn't necessary. The geometry between the lines are so powerful that it naturally doesn't require SET theory. I started by spotting that 5's had to belong on the even line in box 5. That led to 5's in r6c89 and r78c3. This forced dual 5's onto the odd line in box 1. This helped me so much, because it forced extra digits that had to belong in column 3 of box 7. In addition, I had noticed whatever digit accompanied 5 in column 3 of box 7 had to be included in row 6 of box 4, along with a pair that added to 10. All that together led to some tight geometry that allowed me to avoid SET Theory and reduce digits and finally place an 8 into r3c2 and a 9 into r5c2. It is quite amazing that Simon and I had such different styles, each spotting various parts way before the other. What fun!
theres a way to break in without going through cases. it uses the similar sets as simon, except without the orange col 4 but adds in instead that r6c7 is in one of r4c4 or r5c4. we can then virtually cancel them out of both sets. this means there's nothing left in orange, and the top left zipper plus one of r4c4 or r5c4 left in green. since green had one extra set of the digits 1-9, they add to the secret. also all of the greens on the zipper line add to a multiple of 5, and the only one that leave the last green cell as a digit from 1-9 is 8, with the last green = 5.
1:37:50 "Can five be here? I don't think it does anything." I actually facepalmed in real life hard enough for it to hurt.
I could never have spotted the set logic on my own, but after watching this video I did see a slightly different way to look at it.
Highlight boxes 1 and 4 in one color. Then highlight column 3 and row 6 in another. Do the overlapping cancelations, remember that R6C3 is double counted. Then cancel out R5C2 with R6C3 and R9C3. What you end up with is the same set Simon had in green (call it 5y) vs four sets of the other color (call it 4x). So now you know you need a total that is divisible by 4 and 5. In my mind it is a little less convoluted than the set Simon ended up with.
This took me three months to solve. I opened the puzzle in a tab the day this video was published, then I kept looking at it occasionally without figuring out how to break in. I thought about sets, but I just couldn't come up with anything that worked, and especially not something as convoluted as what Simon did here. I ended up focusing on the zipper line that occupies most of box 5 and was able to rule out 11 because that would force the two outer cells on both ends into the upper left zipper and the middle cell of the one below it. Then I checked every possible combination of digits they could be and nothing worked, because it usually resulted in a repeated 9 in row 5. The same logic also ruled out 10 as the sum with no 5 in the outer cells.
I got the 8 in box 1 by figuring out two digits would have to be absent from that zipper, and these digits would have to appear in the three cells below the zipper in box 4, and the only combination of digits that worked was 9 and 4. I had already figured out the upper left zipper would be an 8 or 9, and there was no combination of digits summing to 9 that would be able to repeat below it.
i will stick to my master level sudokus. these insane ones should be called seppuku instead
Wow, what a great evening's entertainment! Simon, your mind is out of this world. And IcyFruit, out of the universe!
so proud to have done this in 86:25! and the break-ins are really thought provoking: it's the importance of numbers 159, what do you put in the cell r6c3, where does 4 go in column 4.
very nice puzzle for such a simple ruleset.
One can deduce from the start that the sum of the line in the center box must be bigger than 9 since 9 in box 5 must be on it.
Brilliant construction and brilliant solve
I wonder if that set logic followed by Simon was the intended "break in" or IcyFruit had some simpler solve path in mind
Either way, finding any path through the puzzle is quite an achievement
A comment on the apps--wish the Steam versions had achievements!
Let us just sit back and enjoy the movie masterpiece from IcyFruit and seeing Simon work his magic in solving this!!
Simon always notices and deduces a bajillion times more than I do, so it's particularly thrilling when I see something he doesn't.
"Where is 5 in column 3?" leads to be bunch of stuff.
He did eventually get to the very similar, "Where is 5 in row 6?" question.
imagine zipper line puzzle where you have incomplete zippers that you have to finish drawing
That actually sounds interesting --- though I think I'll hold out for the Snackdoku version first! 😸
You could have saved 20 minutes at the start just by asking where 9 goes in box 5. It can't go on the short segment, because that line sums to a single digit, so the long segment has a sum of at least 10. It can't be higher than 11, because it has to have four ways to make it, with no doubled digits. In box 9, that line also needs four ways, so it sums to 9, 10, or 11, with the centre being 9, 5, or 1 respectively. That digit must appear in R7/9C4, and therefore must be on the long segment in box 5. This means the line sums are different. R6C7 must appear in R4/5C4 along with the digit which isn't used to make the four pairs, which must be 1 or 5.
I didn't use SET per se, but my head calculations had the same effect as your SET. That allowed me to work out that the long segment in box 5 must have a sum of 10. This is because the three sets of three cells in C4 are some permutation of 12, 15, and 18. The top + bottom + X must be divisible by 5. The only combination which works is 12 + 18 + 10.
@ 1:27:22 - "How am I going to do this?" - Ask where 5 goes in C3. If it's in box 4, it aligns with the 5s in box 5, so pushes 5 onto the 10-line in box 6, where it also needs a 5 on the other end, clashing with the hypothetical 5. Therefore there is a 5 in R7/8C3 and R6C8/9. This forces 5 onto the hook in box 4, which forces 3 into the 3 cells at the other end, and therefore not in R2C2/C3C1, so they are a 17 pair.
@ 1:44:21 - You know 356 are on the line in box 1, so you know 235 are on it in box 4, ruling 36 off the diagonal. This places 7 in R5C3, 6 in R6C3, and 3 in R9C3
@ 1:49:15 - You're worried that you can't recall why you pencil-marked 2s into C1 in box 1 - The whole point of pencil-marking is that once you've made a deduction, you write it down and can forget about how you worked it out, freeing your mind to think about something else. If you're going to mistrust them, there's not really much point in making them. You might as well just work it out from scratch repeatedly.
@ 1:59:48 - "And now if that's a 5" - how can it be? It's part of a 58 pair where the other is in a box with an 8. Don't make rubbish pencil-marks. You work on some logic, and put in pencil-marks without checking whether other rules (like sudoku) resolve them. You waste a lot of time considering stuff that could have been ruled out when you made the pencil-marks.
@ 2:06:40 - "It is 6, 7, or 9" - You've just done it again, making rubbish pencil-marks. There's a 46 pair immediately to the left, it can't be 6.
@ 2:07:37 - "that's 8, or 8 is there" - And again! There's an 8 above your second pencil-mark. You could resolve both the 8 and the 3, if only you looked.
This was quite tough in places, but very interesting. I never got stuck, but there were spells where a sequence of very minor deductions just kept things moving on. Suddenly, it achieved critical mass and gave in.
When you spot something before Simon does and he then spends the next 10 min going over incredibly complex stuff and you're like "you were so close to seeing it!" And then he finds something else that I never would have spotted on my own. lol
I think it is easier to see this puzzle with odd/even polarity. The larger lower line is mod 3 AND mod 4. R1C4 then becomes very constrained, and has a parity affect on C4. Also look at how R5C2 interacts with purple. Near the end, look at how many evens are available in box 2, and what that implies for the length 6 line.
In 15 seconds I got to where Simon got to in 15 minutes, then in the next 2 hrs Simon got to where I will never get to. I'm glad I gave up on this one. I don't mind spending a week on a puzzle. I've done that many times with puzzles on LMG, but I would never have figured this one out.
1:39:00 the 5 can't be in R6C1/C2 because you force the position of 4 and 9 in the same box, making R4C3 impossible (repeat 4 or 5 in the box)
“Gosh, there’s quite a lot of maths here.”
There's a (slightly) easier way to get the first digit. Once you get the SET coloring pattern equating the top left zipper line to most of column 4 + r6c7, You can use your earlier observation that the central zipper line forces the digit from r6c7 into either r4c4 or r5c4, which are the two cells from column 4 not already in the orange set. That means that the green set sums to 45 minus one digit. 40 is the only possible multiple of 5 satisfying that condition, giving the 8 in r3c2.
Finished the puzzle, however needed help with that orange/green set break in. After that the puzzle was no problem.
'I've got the wrong glasses on'. Lol
Yeah. At our age, that can be dangerous. Going down the stairs with readers.
Ya'll stay safe out there :)
Wow. Well, I had to come back to the video a couple times to get a prod; I was looking for the set theory quite a bit early on, and I was in the right general place, but I didn't find the sets --- quite the insight there. And I got stuck later winnowing down r1c4 because I couldn't see the conflict, which was just a matter of not being comfortable with the logic with these lines, I think. A very educational puzzle even in "defeat". 170:49
Holy.... 2+hrs
It's 4.30am... well, I doubt I'll get through it all now, but I very much look forward to finishing it tomorrow.
Congrats on the 575k!!
Whenever I see these videos published, I normally pause the video and then attempt the solve, and either watch the video after I solve it, or let is go on in the background if I get stuck. This puzzle I started the solve, and was immediately attracted to the lines in box-9. I then somewhat quickly determined that the cell in the middle of that box had to be 1, 5 or 9... then started looking for the next path.. but couldn't see it immediately... so I then started looking elsewhere... mostly box-1... and started looking at the options there. Wasn't seeing an easy path there, so I then glanced at the video to see how long the video was... 2 hours?!!! oh boy... if it takes him 2 hours to solve it... will probably take me 10 hours (if I even can...)... so I then decided to just watch the video of Simon solving it (haven't watched it all yet... at 16:32 so far, be interesting to see what amazing tricks needed to solve this).
To add a little more clarity to my statement about "the next path". once I noticed the single cell in box-9 that wasn't on a line, and then noticed there was 2 other boxes that also only had one cell with no line in it (plus box 4 which also caught my attention)... that is what I was looking at guessing that it was the path to follow... but couldn't see anything immediate.
At the 19 minute mark, 9 does not work for the total. Where would you put 9 in the middle box? You can not put it in column 4.
Yep. I spent most of the first 10 minutes of the solve telling the screen "9 has to appear in this box somewhere which means X is at least 10, and r6c7 appears in r4/5c4 so X can't be bigger than 11 or the box total will be over 45!"
Row 6, column 4?
Simon, having watched your geometry analysis I realise there is a far easier way by considering boxes 1 and 4 with column 3 and row 6. Then after canceling the centre digit of box 4 to leave just the “ multiple of 5” zipper as you did, the other set is simply 4 pairs so since the total must exceed 20, the only other multiple of 4x5 is 40 hence the 8 in box 2 and the 10-sum dominoes for box 5. Of course it’s easy to see once you’d suggested the “set” method
Oh I just read Duncan Booth’s comment which says the same
You started trying lines vs boxes but then had to try a lot of cases before reaching a conclusion. However, if you had just added the rest of a box to both sets the answer crystalizes simply. Well, you'd have to know where the furthest right cell is in the other set and the possible values of the final cell, but luckily that I had before I started looking at this. I hope that's clear enough without being too clear.
Had this thought a while ago, but since you mentioned a 600k subscriber celebration, I think the best idea for a special video would be to solve a sudoku while flying with Maverick. Maybe save it for 1mil!
Gotta finish watching this today since I fell asleep last night after only an hour and 10 minutes ( I think Simon got one digit) All I remember is that the madman on the screen started shouting about set theory nobody's ever seen before.
Then I dreamed I went to Surrey for a wedding and we were all waiting on Simon to finish his puzzle before we could start the festivities, but nobody knew when the puzzle would be solved. 😅 Strange things in the world of fever dreams and logic puzzles indeed.
I had exacty 0% chance of finding that wonky set. Great sovle!
1:53:46
what you wanted to say is:
those are the same modular 3
1:46:13 I don't get it. Why Simon doesn't go back to the 5 in box 5 after realising the zipper line centered in box 4 has 9 in the center.
I got pretty quick to the point that the line in box 5 wasn’t a 9 line, as the line in box 4 couldn’t exceed 9, meaning that the box 5 line must have 9 on it (otherwise you would have a 9/0 pair on the box 4 line). Then it was a screaming halt from there…
I think it would have helped Simon a lot to get rid of the colouring once the colouring served its purpose. This would also make it easier to recycle colors elsewhere. It would also help make the grid look nicer, since a lot of the times, the colouring can make the grid look very jarring.
1:38:12 you completely forgot about the 5 forcing the 9 in the middle of box 4 and thus forcing double 5s in the box!
Simon apologizing for the length of a video with an epic solve that most of the audience will have enjoyed immensely is the most Simon thing ever. That and totally overlooking obvious sudoku eliminations making us yell at our screens. 😜
I'm very happy to see this one done by Simon, as I tried it... for hours. And failed it. I found most of the first elements, the 9 10 11 sums, putting 1 5 or 9 in column 4 and preventing the sums to be the same. I found the modulo 3 thing in line 2. But, my god, that SET. I was wandering if there are other ways to disambiguate it all. Was that set really mandatory to get it through ?
We love long videos, please stop apologising!
Love these monstrous puzzles. Nearly six and a half hours for me. But for someone who doesn't like pencil marks cluttering up the grid, there is an awful lot of superfluous colouring in Simon's grid.
Simon not ruling out 9 in the middle box had me briefly upset
I love the fact, that you can not have a 6 on the "bulb" in C4 because you would have a double 2 on the line but can only have an 8 on the "bulb" if you have a double 4.😁
One hour in, and finally a number. Good lord.
1:15:40 why is 2 limited to the 1st column only (in box 4)?
(Edit: The answer is found)
Ah, I've got this:
2 is not allowed in R4C2 because Simon found out earlier that this cell is a repeated one in box 1.
wow I kind of tuned out around 30 minutes. Sometimes I don't mind long puzzles, but this isn't one of them. I came back at 1 hour 30 and......you have 3 digits. lol that's crazy. I basically tuned out because I was completely lost.
I will say that the break-in would be cleaner if you moved the final set of orange cells in column-4 into box-5. Then you are left with 5*green=4*orange.
I feel reasonably clever for having worked out the center line sums to 10 quite quickly, even if the rest of the puzzle took me 4 hours.😅
an easier way of seeing how the zipper in box 5 had to add up to 10 is to note that it needs 5 ways of adding up, as r6c3 will end up seeing all 4 ways of adding to 9 and 11, therefore repeating on its own zipper line in box 5, and you end up with a repeated digit on the zipper line in box 7
Edit: Ignore what I originally wrote. Someone else has written a similar comment, and it has finally sunk in that the cell you are talking about (r6c3) is not r6c4. r6c3 does indeed see four ways of making X. Apologies.
@GiantPeccan replied in another comment that this isn't correct. The pairs summing to X formed by (r6c8, r7c3) and (r6c9, r8c3) could be the same pair, just reversed. So r6c3 doesn't have to be seeing four different ways of making X along row 6 and down column 3.
@@RichSmith77 i see that now, i stand corrected. i was blinded by my own thoughts that id stumbled across simple and applicable logic xD
About 2 hours for me, not sure because I left the timer running for a while. But I cheated a bit. I had decided not to try it and then saw a comment about the set break in giving a 4x=5y equation. I opened the puzzle to see it and the ended up solving it. It was really fun
Im still racking my brain to figure out why C3R1-2 needs to be %3. Wouldnt something like 1-4-7-3-6 have filled out the line at the time of that deduction? I feel like the 2-4 pair there got stumbled on. I would have never been able to come close to solving these, kudos to featuring it. Im just feeling like im missing the logic
Why dose RUclips keep changing the quality to 144p. Makes me wanna eat my hands
I didn't get very far before I decided to just watch SImon do his wizardry, but it did seem to be pretty clear to me from the get-go that the centers of the two lines in boxes 1 and 4 had to be 8 and 9, since by the geometry of the two zipper lines, there needed to be 4 distinct ways of adding pairs up to the center value. Was I making it too easy?
He ends up only using three ways to make the 8 on the box 1 zipper though, since 4+4 is ruled out. And if it's possible in 3 ways, I don't think there's an easy way to rule out 7.
Icy fruit = phistomfel ❤
I never thought Zipper Lines would blow up like this!
One thing I appreciate is that you in the end apologize to people who want to have a more manageable schedule and approachable puzzles instead of monstrous tasks every night.
Houston, we have a problem... Although I do genuinely like the extensive movie-length solve videos very much, I have to confess that it becomes increasingly difficult for me to watch them in practice. I could often afford to do so during lockdown times, but far less so in "normal times", given other responsibilities. If there are too many long videos in one week, I often take my refuge to re-watching older CtC videos, but most of them I know already. ;-) So I am glad lately whenever I spot relatively short new videos. Maybe I'll try to watch the present solve bit by bit in the upcoming week, for it looks quite interesting all the same. Meanwhile, I keep being blown away by all the effort, love and spirit being invested in all these solves by Mark and Simon day after day!
I watch the long videos at 1.75x, and they are still very enjoyable and understandable. Saves some time on the long solves.
@@shzne35Thanks, that is a useful advice I didnt yet think of!
speaking of the app.......I love the latest update where you can apply empty stars. I just wish that feature was on ALL the apps. Or better yet.....Please re-release all the old puzzles as DLC on the new app. I don't mind buying them again on both android and steam. That's fine. I just hate that I don't have the empty stars. Also android has many updates for the old apps (not including empty stars) using the new UI for the CTC app. But steam didn't get those updates. This is why I would love to just have everything on a single app.
Two words: AMAZING
FTR: I could watch a two-hour solve every day; this puzzle is phenomenal
yeah no, completely went over my head this one, i gave up
110:45 for me. Brutal puzzle!
Column 1,2,3 and row 6. Box 1,4,5,7. Set. Yes... That makes row 6 column 789 equal 6 cells in box 5. 3 cells sum to at least 23 . So you can polarity shade.. hang on .. 😮 think im wrong. Sorry 😟