Hi, damasosos92 here! Yep, it's a debut on the channel! ^^ It was just amazing and unexpected to see Glyphs #5 featured here! I know that it's a brutal one, and testers indeed requested me to send a solution guide just to be sure that a logical path existed! You did so well. You just found every intended logic, step after step, and it was so a pleasure to watch. You found literally everything hidden in the puzzle! When you understood the fact that every box has at least three line-cells because of the possible loops in box 3 and box 7 (and so r5c8 is grey) I smiled a lot: that's the part I'm more proud of! The entire puzzle is constructed around the properties of the middle cell of each box given these rules. Around 59:00 the intended idea was to find interactions between boxes 1, 4 and 7, slightly in the same way Simon did. Due to the composition of a 21 cage in three cells (489, 579 or 678), if the middle cell of box 4 was not orange, than the cell above (r4c2) would be orange and it had to connect to box 1, and it's not possible, because the line cells in box 1 are 1234. So the middle cell of box 4 is orange, and so it's a 7. It solves the 3 in box 1, and so on. Oh, about the name of the puzzle: it's the last (and the most difficult one) of a series of sudokus called "Glyphs" in which I used those consecutive line. The name arose because the shapes formed by the lines are like symbols on the grid. I cannot help but invite everyone to try the previous ones. Thank you Simon for the solve and for the kind words, it was really a blast!
I wouldn´t call it "brutal". Somehow I just saw the solutionary numbers in each box. So I had to prove I am right. And after figuring out summing up with just 2 is impossible I only had struggles with box 6. After that it was quite easy - 51 minutes Well done for an debute^^
Funny you should mention job interviews. I've recently concluded a successful job search, and halfway through I added "variant sudoku and Japanese pencil puzzles" to the "Interests" section of my CV. It inevitably cropped up in the subsequent interviews because my interviewers wanted to know what a Japanese pencil puzzle was, and how one might go about varying a sudoku puzzle. Quite helpful for looking like a clever-clogs! I didn't mention that I'm not actually particularly good at either variant sudoku or Japanese pencil puzzles.
😂 For me, the interview would go like this: “I’m really fascinated by extremely complicated sudoku variants and Times Cryptic Crosswords that I can’t get anywhere near solving myself, but I enjoy watching this brilliant British man walk through solving them.” “Ooookay, thank you for your time.”
I like how it used to be a special event for videos to exceed an hour but now it’s becoming more frequent due to new ideas and a massive increase in puzzle logic.
Agree, last year everything over an hour would have been called a movie. Now they're getting close to 2 hours. They might have to go and split their movies soon to make trilogies, lol.
1:26:01 my guy created the most complicated solution when you can deduct that the line cannot go through the middle box from boxes 2 and/or 3 because it would require 4 digits in the middle box in order to pass through and enter the 4th box which is impossible because he already determined the middle box must contain 3 digits.
12:45 Just me or does Simon miss a pencil mark here? I don't see anything we know at this juncture that prevents the 3 from being a diagonal step away from the 2...
(Edit: I'm wrong, see below) Looks like it's provably not possible. if r3c3 is three, the line connects to it. That means we need a 4 in box one that has the line. That means two options. r2c4, r2c3 (which would be four), and r3c4 are all in the line, which makes a 2x2, so this is impossible. Or r4c2, r3c2 (which would be four), and r4c3 are all on the line, which makes a 2x2, so it's impossible. So yeah, at this point you can prove the 3 can't be a diagonal, but he didn't prove it in the video.
@@Mason11987 At this point the 3 in b1 could connect to the 4 in b1. If it did so, the 4 in b1 only needs a single route of escape. Your conclusions are all based on the 4 in b1 being separated from the 3 in b1, which has not been proven.
I've been watching this channel for years, never left a comment. The instant initial 7 blew my mind. You're right of course, and I'd have seen it eventually, but you really just think on a different level. Impressive stuff my man!
The other way to prove the path goes box 1 2 3 6 5 4 is to note that if it does anything else you run out of path cells before you can link all the boxes.
You also need to note that the line can't cross itself. If it could, you might be able to trade a rightward cell in box 5 for a leftward one in box 2, but the rules prohibit it, as well as the fact that crossing would have to be two diagonals forming a 2x2.
Simon got quite lucky that the 3 in box 1 ended up being correct -- at the point that he put it in there was nothing saying it couldn't have been in R3C3. He corner-marked 3 in the box incorrectly very early on in the solve.
He did know that R3C2 needed to be on the line to avoid a grey 2x2 and, since it needed to escape to R2C3 or R3C3, it had to be the 3 in the 1234 sequence that was proven to satisfy the 10 cage.
@Catalynz Yeah I was thinking the same. @scottishrob13 He meant all the way back in the beginning, when Simon placed the pencilamarks for the 3, he forgot to pencilmark r3rc. That could have ruined the whole puzzle.
Well not really, he did have an erroneous pencil mark, but there was enough information available. He already knew that R3C2 was on the line and could only be 3 or 4, and it can't be a 4, because there's nothing for it to connect with.
I am very proud to have finished this in less time than the video, I was surprised I managed so well! Thanks Simon for all the wonderful videos and the great lessons in logic!
I wish Simon had considered the big picture of the possibilities for the overall line sequence, both in terms of consecutive digits and box totals (and overall number of digits, too). There was a lot of restriction from very early on. He'd have gotten box 9 much more quickly, I think. It's easy to see an overall solution, for sure, which doesn't mean it's true, of course. (Although the easy-to-see basic path does turn out to be true.) BUT, it helps you realize where consecutive digit problems are going to exist and allows you to eliminate a lot of red herring paths more quickly and easily. EDIT: I see he finally did look at this around 56:00. He could have used the path-marked 89 in box 6 to progress even earlier. There's only one way that works with boxes 3 and 5, and now it happens to support his overall notion. :) EDIT #2: Around 1:25:00, when Simon is wondering about the case where the path-8 in box 6 connects to a 7 or 9 in box 5, the question to ask is what happens if it's a 9. The next digit on the path will need to be an 8, and that makes this case an impossibility for the cage totals and for overall connectivity. So, it has to be a 7. This must continue in box 5, and the next digit can't be an 8 for the same reason as above, so the next digit is a 6, and then to make box 5 work, the last digit in the box (and the 28th digit on the line) must be a 5.
When I heard the grandfather clock ringing midnight, I suddenly realized that Simon is trying to do this extremely difficult puzzle late at night, probably after doing a couple other puzzles earlier in the day already, which just adds to his mental fatigue. I get frustrated sometimes with him second guessing himself over seemingly obvious things, but if he was as tired as I expect he was, then I can at least understand.
12:50, why is Simon stating that the 3 cannot go diagonal from the 2 in b1? How did he prove this? If this happens, the line in b1 cannot break into two pieces, true, but this isn't a problem at this point. 34:57, again, Simon ignores diagonally connecting the line for no reason.
15:20 What on Earth are you meant to do now? Exactly my sentiments and why I tuned into the video to find out. But I trust you to figure it out and elucidate, whereas I can't even try to work that hard.
Simon too good at sudoku for his own good sometimes. If I were solving this I’d just go “of course it’s 3-4-5 in box 2” and not give it a second thought, and I’d be 100% right in much less time because I’m not clever enough to see a problem with it…
I was completely and utterly nowhere for the first 25 minutes of the puzzle because I forgot the 29 digits limitation. I then finished in 55:33. Didn't even realize it was a 5/5 difficulty. It was probably an advantage not knowing. Once I got the line through boxes 4-9, sudoku became my friend and straightened out the top 3 boxes. Great puzzle!
i had a nice way to show how thing generally worked at around the 1:11:11 mark. we have to connect the whole path, so we know we have to go from box 1 somehow get to box 3, not box 5 since we'd strand the loop in box 3 otherwise then back to box 4. to do so, we need to use at least 6 squares, from boxes 2 and 5. we already knew that each can only have 3 path pieces each, so we need exactly 3. so how high can the bottom path go, i.e. can it get into box 2? the answer is no. otherwise we wouldn't be able to get back to the bottom of box 5 and create a 2x2 of grey. so the bottom path is entirely in box 5 and the upper entirely in box 2. this gets us 3, 4, 5 pencil marks in col 4, 5, 6 of box 2 (resp) and 5, 6, 7 in the cols of box 5. this jumps us to about 1:30:00 in the vid
I never know wether i should think about the puzzle and then being annoyed that he did not see what I was seeing, or not thinking about it and feeling bad. 😂😂
Lately I've been trying to solve the featured puzzles on this channel faster than their time and this is the first time I've beaten it! Not only that but my solve was a staggering 33m18s on this 5/5 difficulty puzzle. I guess sometimes you just find the solve path really quickly. I'll have to watch the whole video now to see what the hang ups were.
I think the way to force the general path Simon conjectured pretty early on and turned out to be correct is to think about how many times you can cross from one side of the grid to the other while only taking the allowed number of cells, from a pretty early point if it doubles back out of box 2 without going straight to box 3 you quickly run out of options to connect all the live ends without using up too many cells in the process...
You must have 3 path cells at least per 3x3 box, otherwise there'll be a 2x2 box that isn't in the path. You have 29 path cells in total, and 27 are required to fill the boxes. That means you only have 2 cells for wiggle room, and you know where they are because you need 1 in the first box (1234) and you need something specific to be happening in the last box otherwise it wouldn't add up to 20. Once you know that every other box has 3 path cells it becomes very easy to fill the line.
Agree - Simon unfortunately missed this rather obvious property and spent a lot of time struggling in the middle of the video as a result. I felt bad for him as the logic is painfully difficult if this isn’t concluded
How did I solve this puzzle? Well, I actually chose to solve a slightly different puzzle from this one, the puzzle of where I was going to find over 100 minutes to watch you solve so interesting a sudoku. I solved that puzzle by watching in two sessions - and believe me, I was very eager for my "intermission" to end so that I could watch the second half. On this video, as on so many others of the very long ones, I think that, in addition to "brilliant" and "clever" and "genius" one must add "stubborn" as an attribute - you would not give up, which is wonderful. Thanks, Simon, for burning the midnight oil to create this video for us.
33:00 for me. When you realise early that boxes 6,8 and 9 must be combined from separate entries (not mod 3) the in and out makes the flow clear and it allows us to work out the flow of boxes and then due to consecutive it has a particular order. Thanks so much for this puzzle
The only reason why it took him so long is that he didn't realize he had to put at least three cell paths in every 3x3 box. When you realize that it becomes extremely easy to solve because you know you don't have enough wiggle room to jump back from box to box, as the first and last box must be the ones with 4 cells.
Took me 126:53 which I'm very proud of. After doing my solve, and then watching Simon's solve, I realised I probably made some assumptions that just turned out to be correct. For example, I worked out that box 1 had to hit box 2 and I assumed that it went 1234, then 345 in order to satisfy the box numbers so was a bit restricted. Looking at Simon questioning that path made me realise I hadn't considered some of the other options that Simon then had to invalidate. Very clever puzzle!! Thankyou.
Wow, what a puzzle. I managed to solve it just under an hour. Something about this puzzle just spoke to me and it looks like I was able to solve it faster than most. My big breakthrough was discovering the aside from box 1 and 9, each other box had to have 3 cells on the line as both the starting and end points necessitated 4 cells on the line in their boxes due to the numbers and how they summed. Great puzzle and the fun will be watching Simon solve it and seeing if we had similar solve paths.
One small tip that may help solving this faster is noticing that once ya got r4c8 being part of the line and being 89 is that that cell is the connection to block 3, meaning that block 6 can only connect to block 5 and block 3 only to block 2 (math doesnt add up otherwise if you try to do another in&out with either or those to connect to other block) meaning block 2 connects to block 1 and block 5 to 4 (since they are consecutive numbers) meaning the rest of the grid becomes normal sudoku
I think Simon would've been quicker if he considered the consecutive digit rule more between the cages. Once you look at box 1 and 4 for instance, you'll realize immediately you can't go into r4c2, as you can do 5/6 at most coming from box 1 and you need another 10. Difficult to pinpoint how much time it would save but it would definitely get him unstuck more often.
Loved this one. Took me roughly 2 hours to get it down, which I'm calling a win, hahaha... Now watching the solve, and I find it interested that Simon never seems to take the "graphical" route. The first thing I did on the puzzle was to plot the most direct line possible, and then see where deviations were needed to satisfy the rules, most importantly the 29 cell rule, the no 2x2 rule, and the consecutive digit rule, which I first had misread as to mean that the line has to be completely sequential (as in, 1-9 and then 9-1, etc)... I had the line down remarkably close to how it ended before placing as many as three digits in the grid. Had a lot of trouble in box 2, and was stuck there for almost an entire hour trying to figure out the exact arrangement of the digits, but it worked out well enough in the end. Wonderful puzzle.
35:53 for me and it usually takes me ages. Came out really easily for me. I started by working out what the sequence had to be in each cage and the line after that was quite restricted.
Having established that every box can only have 3 numbers on the path I was amazed how long it took to realise that the 3,4,5 across box 2 had to be left to right - there's no other way to make the path!
Deserved 5 / 100 for the puzzle and 100 for Simon. I could see some pieces of solution, but was too lazy (or just chicken) to go the entire way through. Thank You, Simon! Amazing solution and comments.
very strange puzzle. i stared at it for an hour and only got as far as knowing how many line cells were in each box, but couldn't figure out how to definitively prove where it actually went. so i winged it based on what i thought the solution /should/ be, and solved the whole thing in 15 minutes from there (specifically my train of thought was "there can't be /too/ many places where the line leaves and re-enters a box", and it was conspicuous that a lot of boxes could be expressed as consecutive digits, so i just drew a line that used as many of those as possible, then nudged it around a bit to fix rule violations)
r2c8 has to be orange to avoid grey 2x2... but there can only be 3 oranges (to avoid orange 2x2)... therefore 3 consecutive cells adding ro 21 has a middle cell of 7 with 6 8 either side
I've massive upgraded from numpty to decent toiler (rather than superstar ) over the past few years with CtC. 70 minutes for me on this, which I'm quite chuffed with. Even with forgetting the no border cells for a bit!
58:00 That actually doesn't work because r2c3 would be the end of the 1234 sequence and r2c5 would be the middle of the 345 / 543 sequence, making them both 4s even though they're in the same row.
As a non pro and as someone who normally takes 8-10 hours to solve a puzzle that takes Simon 1 hour.... I finally matched his time, just about. 97:58, here.
One of the VERY rare occasions I solved faster than Simon, although to be fair I wasn't trying it at midnight! The 3,4,5 across box 2 going left to right to join to box 3 was evident for a long time (once box 1 to box 4 was ruled out at @1:04:10 ); there's no way that box 5 could join to box 3 and get back to the path in box 4 with only 3 numbers on the path in each box.
62:34 For me, my night brain was not having it but my morning in-work brain couldn't help but think about this and make quite a breakthrough soon after
This one took me forever; I kind of did accidental bifurcation because I had a couple deductions that I THOUGHT were forced and they turned out to be wrong and I had to start over. But I eventually landed on the correct logical path.
for anyone wondering *blank* mod *blank*, the mod is the number, 22/18/9/12 divided by how many cells he needs in each box ie 3. the mod is the remainder, which is why 22/3 has a mod of 1 *I think*. if I got that wrong let me know 😂
After getting r7c7 orange simon just colors r8c7 orange aswell, but couldnt r7c7 be a single 5 that gets poked in from outside box 9? Or am i just missing something here?
Heyyy guys! Big ups, can you do a setting video at some point again? Maybe collab or something with Phistomefel... I'm trying to get into setting this year and hopefully competing as well in the NZ National Sudoku Champs... much love from New Zealand - Cass
1:28:51, but with an asterisk as (1) I somewhat guessed my way through after a slow start (but visually what the path 'ought' to look like), and (2) at one point I got it in my head that r2c5 was on the path, which led to a break and me taking a quickie peek late in the Simon solution to see where I erred. I don't care, I'm counting this as at least a partial victory. Tough puzzle!
Interesting! Quite curious to see whether daytime Simon would have been able to see a bit more high level logic on the trajectory of the line as mentioned in other comments here, rather than midnight Simon struggling to find deductions on this, and how much shorter the video would be.
I think there was an efficient path requirement in this puzzle that Simon didn't talk about. The path couldn't meander very much within the restriction of the 3 average per box. I may have missed it since it's a long video though.
9 boxes with sums. Only the 9, 10, 12 could possibly be 2 digits. The 10 is forced to 1234 immediately. The 9 & 12 with 2 digits would "strand" a 2x2 without line. The 20 must also be 4 digits. (A single 5 in the top left would strand a 2x2, a lower 5 would need to connect to 4 & 6 in the 9 cage, breaking it.) 7 x 3 + 4 + 4 = 29 That's all 29 line segments accounted for. The rest must all be 3 cell sums.
Wow, this was the first puzzle I've done faster than Simon. And when I saw how long the video was I thought to myself, I might as well not even try it. I do think I got a little lucky early on though. I think I might have made an assumption that turned out to be true but didn't need to be.
Lol. Funny one Go to sleep, Simon. [Yeah, count the sheep in your head -- or floating above your head -- going in circles -- which is what you had me doing with that "cursor" sitting on r6c6 @ 1:33:08
48:55 with a couple of dumb mistakes. I thought I ruined Box 6, forgetting I could enter it from Box 9 and then go right back out. I messed up the 4 in Box 1, forgetting I could move in a diaganol at the very beginning. And when checking my mistakes, I realized that I assumed the 6 in Box 4 was in Row 4 without considering it being in Row 6. The line moving in a diaganol kept messing me up. I also did this kinda fast on my phone, just going with my logic, not taking the time to prove it 100%, which ended up being the problem, haha.
I know in this puzzle the rules say sequential upon the line, but in other puzzles you’ve made this same deduction of the x-1+ x + x+2 = 3x and left it at that? So how come you’ve never done x-2 + x + x+2 = 3x. So in other scenarios it could be 579. As x -2 +x + x+2 still share the same properties and = 3x. If the digits on either side of the central digit are the same distance away from it, they will always sum to the center digit x3. Taking that to the extreme, 1,5,9 sum to 15 as 4,5,6 do. We can go even further, 1 + 20 + 39 = 60 as does 19,20,21.
I only remember him doing it with three cell renbans (x-1, x, x+1), which also have to be sequential, just they can be in any order, so you don't know x is in the middle of the renban.
You know Simon's getting a bit fatigued when he starts providing explanations like "I'm going to put a pencil mark in there just because... I want to." (1:26:30)
To my shame, I ran out of patience after an hour of looking for a logical path, and found the solution by conjecture (trial and error). Must do better. Simon did a heck of a lot better than I did at proving his solution logically, but did he have a small miss-step at 1:05:00? He seemed to overlook the possibility that the line from box 7 could have escaped to box 5 rather than box 4. I have a feeling the three orange cell limit in box 5 means you run into difficulties if you try to go from box 7 to box 5, and still connect a path via every box. But it's not simple to prove.
My idea when I setted the puzzle was that you prove that the middle cell in box 4 is orange, so it's a 7, and then r6c2 is orange, consecutive to 7 and different to 6, so it's an 8. With this information, that 8 has to connect to something somewhere and the only available cell is the 7 in box 7. If you want to see the simplest intended path I wrote a solution guide at this link: drive.google.com/file/d/1_dD8AX1iGVZxvnpWK5G6xlyGsl5GSrKP/view?usp=sharing ^^
(Un)fortunately, Simon forgetting about the possibility of diagonal lines doesn't result in error😅 Also (un)fortunately, Simon forgets about the 29 cell restriction or doesn't understand how powerful it is
Lol. You cracked me up, Simon. "Can we see any naughty 2×2s?" (@1:33:07, btw) And his cursor is sitting on one (r6c6). What does Simon do?! Alights on one up in row4. Hilarious, Simon. [I knew that center one had to be the 6 though] Still funny, you alighting on another square when your cursor is sitting on a *definite* one. Lol 👍👍😂😎☕️
SIMON, you inadvertently tricked me with your cursor sittin' on r6c6 [And I thought that was a "2×2"] Lol. I'll figure it out from here. Thanks. [I kept getting a conflict near the end] I got tricked by Simon (and his cursor) sitting on a crossroads. [And, boy, did I take the wrogn path, lol] Fool by the "cursor" though, lol. Good one, Simon (even though you didn't know it). 😂😎☕️☕️[
As soon as you know the remaining boxes only have 3 orange cells each you can conclude that the central cell of the grid is orange, because anything else would leave a 2×2 of grey in the central box. Edit: nevermind, I'm wrong.
A little surprised Simon didn't see earlier the jump from box 9 to the 5 in box 6 and then back in to create the 7643 20 total, and then the 234 to make up the 9 total in box 8.
Hi, damasosos92 here!
Yep, it's a debut on the channel! ^^
It was just amazing and unexpected to see Glyphs #5 featured here! I know that it's a brutal one, and testers indeed requested me to send a solution guide just to be sure that a logical path existed! You did so well. You just found every intended logic, step after step, and it was so a pleasure to watch. You found literally everything hidden in the puzzle!
When you understood the fact that every box has at least three line-cells because of the possible loops in box 3 and box 7 (and so r5c8 is grey) I smiled a lot: that's the part I'm more proud of! The entire puzzle is constructed around the properties of the middle cell of each box given these rules.
Around 59:00 the intended idea was to find interactions between boxes 1, 4 and 7, slightly in the same way Simon did. Due to the composition of a 21 cage in three cells (489, 579 or 678), if the middle cell of box 4 was not orange, than the cell above (r4c2) would be orange and it had to connect to box 1, and it's not possible, because the line cells in box 1 are 1234. So the middle cell of box 4 is orange, and so it's a 7. It solves the 3 in box 1, and so on.
Oh, about the name of the puzzle: it's the last (and the most difficult one) of a series of sudokus called "Glyphs" in which I used those consecutive line. The name arose because the shapes formed by the lines are like symbols on the grid. I cannot help but invite everyone to try the previous ones.
Thank you Simon for the solve and for the kind words, it was really a blast!
Congrats on the Feature! Beautiful Puzzle!
Just astonishing debut from you!! Loved every second of it. Congrats!!
Great puzzle! I really enjoyed the logic, especially because I managed to understand it :)
Thank you for the great puzzle!
I wouldn´t call it "brutal". Somehow I just saw the solutionary numbers in each box. So I had to prove I am right. And after figuring out summing up with just 2 is impossible I only had struggles with box 6. After that it was quite easy - 51 minutes
Well done for an debute^^
Funny you should mention job interviews. I've recently concluded a successful job search, and halfway through I added "variant sudoku and Japanese pencil puzzles" to the "Interests" section of my CV. It inevitably cropped up in the subsequent interviews because my interviewers wanted to know what a Japanese pencil puzzle was, and how one might go about varying a sudoku puzzle. Quite helpful for looking like a clever-clogs!
I didn't mention that I'm not actually particularly good at either variant sudoku or Japanese pencil puzzles.
Congratulations on your job search success!
@@emilywilliams3237 Thank you very much!
😂 For me, the interview would go like this: “I’m really fascinated by extremely complicated sudoku variants and Times Cryptic Crosswords that I can’t get anywhere near solving myself, but I enjoy watching this brilliant British man walk through solving them.”
“Ooookay, thank you for your time.”
Soon, we shall have a Cracking the Cryptic - Job search section where we connect employers and job seekers.
I am absolutely going to use it in my next job search
I like how it used to be a special event for videos to exceed an hour but now it’s becoming more frequent due to new ideas and a massive increase in puzzle logic.
Agree, last year everything over an hour would have been called a movie. Now they're getting close to 2 hours. They might have to go and split their movies soon to make trilogies, lol.
@@tabularasa0606 still movies to this time for me and popcorn is a must (yey)
I'm ok with it. I thrive on it
1:26:01 my guy created the most complicated solution when you can deduct that the line cannot go through the middle box from boxes 2 and/or 3 because it would require 4 digits in the middle box in order to pass through and enter the 4th box which is impossible because he already determined the middle box must contain 3 digits.
12:45 Just me or does Simon miss a pencil mark here? I don't see anything we know at this juncture that prevents the 3 from being a diagonal step away from the 2...
(Edit: I'm wrong, see below)
Looks like it's provably not possible.
if r3c3 is three, the line connects to it.
That means we need a 4 in box one that has the line.
That means two options.
r2c4, r2c3 (which would be four), and r3c4 are all in the line, which makes a 2x2, so this is impossible.
Or
r4c2, r3c2 (which would be four), and r4c3 are all on the line, which makes a 2x2, so it's impossible.
So yeah, at this point you can prove the 3 can't be a diagonal, but he didn't prove it in the video.
@@Mason11987 At this point the 3 in b1 could connect to the 4 in b1. If it did so, the 4 in b1 only needs a single route of escape. Your conclusions are all based on the 4 in b1 being separated from the 3 in b1, which has not been proven.
@@RoderickEtheria Ah you're right! I stand corrected
@@Mason11987 Well done.
Correct…
Simon apologizing for being slow on a rough bit of logic (while explaining everything he was thinking about) is pure Simon.
Nice puzzle. Nice solve.
1:29:59 What would Poirot say about "isolating little grey cells in a useful manner"? 😀
I've been watching this channel for years, never left a comment.
The instant initial 7 blew my mind. You're right of course, and I'd have seen it eventually, but you really just think on a different level. Impressive stuff my man!
Rules: 04:10
Let's Get Cracking: 07:31
Simon's time: 1h32m19s
Puzzle Solved: 1:39:50
What about this video's Top Tier Simarkisms?!
Bobbins: 2x (48:01, 1:26:57)
Three In the Corner: 2x (1:39:02, 1:39:27)
Chocolate Teapot: 2x (53:08, 1:35:33)
The Secret: 2x (06:27, 1:29:37)
Scooby-Doo: 1x (55:50)
And how about this video's Simarkisms?!
Hang On: 16x (15:54, 20:21, 28:48, 28:55, 34:17, 43:24, 43:31, 43:31, 43:31, 45:04, 45:32, 48:37, 1:06:47, 1:19:08, 1:19:37, 1:35:24)
Wow: 16x (05:39, 14:41, 14:49, 21:15, 21:15, 21:15, 21:15, 21:15, 21:15, 23:54, 25:09, 50:49, 1:05:15, 1:24:30, 1:38:23, 1:38:23)
Ah: 16x (05:39, 07:58, 13:36, 16:51, 22:52, 24:19, 31:36, 32:31, 40:15, 48:01, 49:07, 1:00:29, 1:05:24, 1:28:03, 1:33:36, 1:35:49)
Sorry: 14x (03:37, 21:53, 23:43, 23:43, 24:50, 32:34, 39:56, 51:08, 56:09, 58:16, 1:00:09, 1:10:01, 1:19:06, 1:22:14)
By Sudoku: 9x (50:58, 1:23:24, 1:23:26, 1:25:49, 1:27:53, 1:33:51, 1:35:49, 1:35:52, 1:36:25)
Clever: 8x (18:45, 21:38, 44:26, 48:12, 55:52, 58:46, 58:49, 58:53)
In Fact: 5x (08:31, 29:32, 34:17, 1:07:28, 1:10:09)
Pencil Mark/mark: 5x (56:29, 56:31, 56:31, 1:26:29, 1:34:38)
Corollary: 4x (08:01, 10:04, 11:57, 37:17)
Obviously: 4x (30:18, 39:33, 1:15:48, 1:40:42)
Brilliant: 3x (1:41:11, 1:41:12, 1:41:33)
Shouting: 3x (02:16, 02:22, 03:58)
Good Grief: 2x (51:05, 1:07:46)
What on Earth: 2x (1:02:08, 1:29:26)
Naked Single: 2x (1:15:03, 1:37:56)
Naughty: 2x (1:33:07, 1:35:49)
I Have no Clue: 2x (1:09:00, 1:09:02)
Beautiful: 2x (01:08, 1:40:10)
Unbelievable: 2x (1:06:42, 1:30:39)
Useless: 1x (1:26:19)
Goodness: 1x (44:00)
Bother: 1x (27:26)
Axiomatically: 1x (07:53)
The Answer is: 1x (13:43)
Out of Nowhere: 1x (53:05)
Nonsense: 1x (1:02:02)
Recalcitrant: 1x (1:29:31)
Bingo: 1x (1:33:20)
Stuck: 1x (1:40:20)
Fascinating: 1x (44:29)
Come on Simon: 1x (22:37)
Approachable: 1x (02:48)
Intriguing: 1x (01:20)
Progress: 1x (15:21)
Fabulous: 1x (04:02)
That's Huge: 1x (42:03)
Have a Think: 1x (1:12:12)
Unique: 1x (1:34:24)
Most popular number(>9), digit and colour this video:
Twenty Two (10 mentions)
Three (164 mentions)
Orange (91 mentions)
Antithesis Battles:
High (2) - Low (0)
Even (12) - Odd (2)
Outside (3) - Inside (0)
Row (5) - Column (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!
Orange is a number?
Orange is a color. As mentioned if you had read it correctly.
@@aloi4 The title says "Most popular number(>9), digit and colour this video:"
Colour is also mentioned in the title.
@@abdullahmansoor1 True lol!
I just understand now
Life saving comment my friend haha thanks bud👌🤣
The other way to prove the path goes box 1 2 3 6 5 4 is to note that if it does anything else you run out of path cells before you can link all the boxes.
You also need to note that the line can't cross itself. If it could, you might be able to trade a rightward cell in box 5 for a leftward one in box 2, but the rules prohibit it, as well as the fact that crossing would have to be two diagonals forming a 2x2.
Simon got quite lucky that the 3 in box 1 ended up being correct -- at the point that he put it in there was nothing saying it couldn't have been in R3C3. He corner-marked 3 in the box incorrectly very early on in the solve.
He did know that R3C2 needed to be on the line to avoid a grey 2x2 and, since it needed to escape to R2C3 or R3C3, it had to be the 3 in the 1234 sequence that was proven to satisfy the 10 cage.
@Catalynz Yeah I was thinking the same. @scottishrob13 He meant all the way back in the beginning, when Simon placed the pencilamarks for the 3, he forgot to pencilmark r3rc. That could have ruined the whole puzzle.
Yeah I thought the same. I guess it was only a 33% chance of being wrong at the time. 12:44 for those curious.
Well not really, he did have an erroneous pencil mark, but there was enough information available. He already knew that R3C2 was on the line and could only be 3 or 4, and it can't be a 4, because there's nothing for it to connect with.
I am very proud to have finished this in less time than the video, I was surprised I managed so well! Thanks Simon for all the wonderful videos and the great lessons in logic!
12:45 couldn't there be a 3 in r3c3?
that is what I was thinking...
I wish Simon had considered the big picture of the possibilities for the overall line sequence, both in terms of consecutive digits and box totals (and overall number of digits, too). There was a lot of restriction from very early on. He'd have gotten box 9 much more quickly, I think. It's easy to see an overall solution, for sure, which doesn't mean it's true, of course. (Although the easy-to-see basic path does turn out to be true.) BUT, it helps you realize where consecutive digit problems are going to exist and allows you to eliminate a lot of red herring paths more quickly and easily.
EDIT: I see he finally did look at this around 56:00. He could have used the path-marked 89 in box 6 to progress even earlier. There's only one way that works with boxes 3 and 5, and now it happens to support his overall notion. :)
EDIT #2: Around 1:25:00, when Simon is wondering about the case where the path-8 in box 6 connects to a 7 or 9 in box 5, the question to ask is what happens if it's a 9. The next digit on the path will need to be an 8, and that makes this case an impossibility for the cage totals and for overall connectivity. So, it has to be a 7. This must continue in box 5, and the next digit can't be an 8 for the same reason as above, so the next digit is a 6, and then to make box 5 work, the last digit in the box (and the 28th digit on the line) must be a 5.
Great movie to start the week going. Thank you Simon for lovely solve and being so thoroughly engaged with it as usual.
When I heard the grandfather clock ringing midnight, I suddenly realized that Simon is trying to do this extremely difficult puzzle late at night, probably after doing a couple other puzzles earlier in the day already, which just adds to his mental fatigue. I get frustrated sometimes with him second guessing himself over seemingly obvious things, but if he was as tired as I expect he was, then I can at least understand.
12:50, why is Simon stating that the 3 cannot go diagonal from the 2 in b1? How did he prove this? If this happens, the line in b1 cannot break into two pieces, true, but this isn't a problem at this point.
34:57, again, Simon ignores diagonally connecting the line for no reason.
At 1:04:53 he also makes an invalid deduction by not considering entering box 5 diagonally.
15:20 What on Earth are you meant to do now? Exactly my sentiments and why I tuned into the video to find out. But I trust you to figure it out and elucidate, whereas I can't even try to work that hard.
Simon too good at sudoku for his own good sometimes. If I were solving this I’d just go “of course it’s 3-4-5 in box 2” and not give it a second thought, and I’d be 100% right in much less time because I’m not clever enough to see a problem with it…
I was completely and utterly nowhere for the first 25 minutes of the puzzle because I forgot the 29 digits limitation. I then finished in 55:33. Didn't even realize it was a 5/5 difficulty. It was probably an advantage not knowing. Once I got the line through boxes 4-9, sudoku became my friend and straightened out the top 3 boxes. Great puzzle!
i had a nice way to show how thing generally worked at around the 1:11:11 mark. we have to connect the whole path, so we know we have to go from box 1 somehow get to box 3, not box 5 since we'd strand the loop in box 3 otherwise then back to box 4. to do so, we need to use at least 6 squares, from boxes 2 and 5. we already knew that each can only have 3 path pieces each, so we need exactly 3. so how high can the bottom path go, i.e. can it get into box 2? the answer is no. otherwise we wouldn't be able to get back to the bottom of box 5 and create a 2x2 of grey. so the bottom path is entirely in box 5 and the upper entirely in box 2. this gets us 3, 4, 5 pencil marks in col 4, 5, 6 of box 2 (resp) and 5, 6, 7 in the cols of box 5. this jumps us to about 1:30:00 in the vid
I never know wether i should think about the puzzle and then being annoyed that he did not see what I was seeing, or not thinking about it and feeling bad. 😂😂
Amazing solve Simon. What a brutal puzzle , I never stood a chance but loved your solve. Thank you.
Lately I've been trying to solve the featured puzzles on this channel faster than their time and this is the first time I've beaten it! Not only that but my solve was a staggering 33m18s on this 5/5 difficulty puzzle. I guess sometimes you just find the solve path really quickly. I'll have to watch the whole video now to see what the hang ups were.
Also found this fairly easy to be honest
Excellent solve Simon. Another brilliantly conceived puzzle.
I think the way to force the general path Simon conjectured pretty early on and turned out to be correct is to think about how many times you can cross from one side of the grid to the other while only taking the allowed number of cells, from a pretty early point if it doubles back out of box 2 without going straight to box 3 you quickly run out of options to connect all the live ends without using up too many cells in the process...
You must have 3 path cells at least per 3x3 box, otherwise there'll be a 2x2 box that isn't in the path. You have 29 path cells in total, and 27 are required to fill the boxes. That means you only have 2 cells for wiggle room, and you know where they are because you need 1 in the first box (1234) and you need something specific to be happening in the last box otherwise it wouldn't add up to 20. Once you know that every other box has 3 path cells it becomes very easy to fill the line.
Agree - Simon unfortunately missed this rather obvious property and spent a lot of time struggling in the middle of the video as a result. I felt bad for him as the logic is painfully difficult if this isn’t concluded
Get the popcorn! Settle down and prepare for an epic!🎉
How did I solve this puzzle? Well, I actually chose to solve a slightly different puzzle from this one, the puzzle of where I was going to find over 100 minutes to watch you solve so interesting a sudoku. I solved that puzzle by watching in two sessions - and believe me, I was very eager for my "intermission" to end so that I could watch the second half. On this video, as on so many others of the very long ones, I think that, in addition to "brilliant" and "clever" and "genius" one must add "stubborn" as an attribute - you would not give up, which is wonderful. Thanks, Simon, for burning the midnight oil to create this video for us.
At 1:03:34 roughly, you draw a black vertical line between boxes 4 and 5 and by the end of the puzzle you never remove it and i can't unsee it :c
And now the rest of us can’t either 😭
@@RhiannonAgutter Misery loves company
33:00 for me. When you realise early that boxes 6,8 and 9 must be combined from separate entries (not mod 3) the in and out makes the flow clear and it allows us to work out the flow of boxes and then due to consecutive it has a particular order. Thanks so much for this puzzle
The only reason why it took him so long is that he didn't realize he had to put at least three cell paths in every 3x3 box. When you realize that it becomes extremely easy to solve because you know you don't have enough wiggle room to jump back from box to box, as the first and last box must be the ones with 4 cells.
Took me 126:53 which I'm very proud of. After doing my solve, and then watching Simon's solve, I realised I probably made some assumptions that just turned out to be correct. For example, I worked out that box 1 had to hit box 2 and I assumed that it went 1234, then 345 in order to satisfy the box numbers so was a bit restricted. Looking at Simon questioning that path made me realise I hadn't considered some of the other options that Simon then had to invalidate. Very clever puzzle!! Thankyou.
Wow, what a puzzle. I managed to solve it just under an hour. Something about this puzzle just spoke to me and it looks like I was able to solve it faster than most. My big breakthrough was discovering the aside from box 1 and 9, each other box had to have 3 cells on the line as both the starting and end points necessitated 4 cells on the line in their boxes due to the numbers and how they summed. Great puzzle and the fun will be watching Simon solve it and seeing if we had similar solve paths.
"It's not huge, but it's profound" - desperate attempt to salvage a date which was going really well up to that point.
One small tip that may help solving this faster is noticing that once ya got r4c8 being part of the line and being 89 is that that cell is the connection to block 3, meaning that block 6 can only connect to block 5 and block 3 only to block 2 (math doesnt add up otherwise if you try to do another in&out with either or those to connect to other block) meaning block 2 connects to block 1 and block 5 to 4 (since they are consecutive numbers) meaning the rest of the grid becomes normal sudoku
I think Simon would've been quicker if he considered the consecutive digit rule more between the cages. Once you look at box 1 and 4 for instance, you'll realize immediately you can't go into r4c2, as you can do 5/6 at most coming from box 1 and you need another 10.
Difficult to pinpoint how much time it would save but it would definitely get him unstuck more often.
Loved this one. Took me roughly 2 hours to get it down, which I'm calling a win, hahaha... Now watching the solve, and I find it interested that Simon never seems to take the "graphical" route. The first thing I did on the puzzle was to plot the most direct line possible, and then see where deviations were needed to satisfy the rules, most importantly the 29 cell rule, the no 2x2 rule, and the consecutive digit rule, which I first had misread as to mean that the line has to be completely sequential (as in, 1-9 and then 9-1, etc)... I had the line down remarkably close to how it ended before placing as many as three digits in the grid. Had a lot of trouble in box 2, and was stuck there for almost an entire hour trying to figure out the exact arrangement of the digits, but it worked out well enough in the end. Wonderful puzzle.
12:58 can someone explain why 3 was ruled out in r3c3 cell?
35:53 for me and it usually takes me ages. Came out really easily for me. I started by working out what the sequence had to be in each cage and the line after that was quite restricted.
Having established that every box can only have 3 numbers on the path I was amazed how long it took to realise that the 3,4,5 across box 2 had to be left to right - there's no other way to make the path!
55:19 for me. What a brutal puzzle!! I really struggled with this one, but I'm very happy I managed to solve it in the end!!
Deserved 5 / 100 for the puzzle and 100 for Simon.
I could see some pieces of solution, but was too lazy (or just chicken) to go the entire way through. Thank You, Simon! Amazing solution and comments.
very strange puzzle. i stared at it for an hour and only got as far as knowing how many line cells were in each box, but couldn't figure out how to definitively prove where it actually went. so i winged it based on what i thought the solution /should/ be, and solved the whole thing in 15 minutes from there
(specifically my train of thought was "there can't be /too/ many places where the line leaves and re-enters a box", and it was conspicuous that a lot of boxes could be expressed as consecutive digits, so i just drew a line that used as many of those as possible, then nudged it around a bit to fix rule violations)
Nice video, although I was very puzzled by the fact that Simon immediately knew that box r8c2 and r2c8 where 6 and 7 respectively.
r2c8 has to be orange to avoid grey 2x2... but there can only be 3 oranges (to avoid orange 2x2)... therefore 3 consecutive cells adding ro 21 has a middle cell of 7 with 6 8 either side
@@gatlygat I honestly had no idea, in fact - I had to use a calculator to figure out why....
I've massive upgraded from numpty to decent toiler (rather than superstar ) over the past few years with CtC. 70 minutes for me on this, which I'm quite chuffed with. Even with forgetting the no border cells for a bit!
58:00 That actually doesn't work because r2c3 would be the end of the 1234 sequence and r2c5 would be the middle of the 345 / 543 sequence, making them both 4s even though they're in the same row.
Thinking to myself "this would be far easier if we assumed uniqueness" only to have Simon say much the same thing in his closing remarks.
As a non pro and as someone who normally takes 8-10 hours to solve a puzzle that takes Simon 1 hour.... I finally matched his time, just about. 97:58, here.
1:15:22 Yay! I beat Simon again! Very enjoyable and original puzzle.
One of the VERY rare occasions I solved faster than Simon, although to be fair I wasn't trying it at midnight!
The 3,4,5 across box 2 going left to right to join to box 3 was evident for a long time (once box 1 to box 4 was ruled out at @1:04:10 ); there's no way that box 5 could join to box 3 and get back to the path in box 4 with only 3 numbers on the path in each box.
62:34 For me, my night brain was not having it but my morning in-work brain couldn't help but think about this and make quite a breakthrough soon after
This one took me forever; I kind of did accidental bifurcation because I had a couple deductions that I THOUGHT were forced and they turned out to be wrong and I had to start over. But I eventually landed on the correct logical path.
"We should basically hire you" or "This person is likely to spend 1h42 in the bogs doing sudoku" :)
45:01 It took me longer because I forgot the "not on the edge" rule. I thought the solution was ambiguous until I remembered that rule.
“That is why it’s got 5 stars out of 5 for difficulty is it’s extremely challenging.”
Great solve, great puzzle.
for anyone wondering *blank* mod *blank*, the mod is the number, 22/18/9/12 divided by how many cells he needs in each box ie 3. the mod is the remainder, which is why 22/3 has a mod of 1 *I think*. if I got that wrong let me know 😂
After getting r7c7 orange simon just colors r8c7 orange aswell, but couldnt r7c7 be a single 5 that gets poked in from outside box 9? Or am i just missing something here?
If R8C7 is grey, then we will have 2*2 grey cells
I don't know how I figured out the logic & solved it faster than Simon, but I did. 1:11:48. I'm astounded I visualized that path correctly.
Me too - in just a couple of minutes under your time. I typically take about 50% longer than Simon to complete a puzzle so this was a surprise.
Heyyy guys! Big ups, can you do a setting video at some point again? Maybe collab or something with Phistomefel... I'm trying to get into setting this year and hopefully competing as well in the NZ National Sudoku Champs... much love from New Zealand
- Cass
Come join the discord server, there are plenty of helpful people there. :)
1:28:51, but with an asterisk as (1) I somewhat guessed my way through after a slow start (but visually what the path 'ought' to look like), and (2) at one point I got it in my head that r2c5 was on the path, which led to a break and me taking a quickie peek late in the Simon solution to see where I erred. I don't care, I'm counting this as at least a partial victory.
Tough puzzle!
30:25 for me. For some reason I had no problem to find the correct path and from there it was very simple...
Interesting! Quite curious to see whether daytime Simon would have been able to see a bit more high level logic on the trajectory of the line as mentioned in other comments here, rather than midnight Simon struggling to find deductions on this, and how much shorter the video would be.
I think there was an efficient path requirement in this puzzle that Simon didn't talk about. The path couldn't meander very much within the restriction of the 3 average per box. I may have missed it since it's a long video though.
I've seen this video like 8 times and I'm just now realizing what the video title means
13:30 why can't 3 be in r3c3?
Simon does love to say Sorry for being slow and your site how is he figuring this out.
9 boxes with sums. Only the 9, 10, 12 could possibly be 2 digits.
The 10 is forced to 1234 immediately.
The 9 & 12 with 2 digits would "strand" a 2x2 without line.
The 20 must also be 4 digits. (A single 5 in the top left would strand a 2x2, a lower 5 would need to connect to 4 & 6 in the 9 cage, breaking it.)
7 x 3 + 4 + 4 = 29
That's all 29 line segments accounted for.
The rest must all be 3 cell sums.
Well, almost. I missed some logic on the 20 combinations.
It seems that intuitive solution might be x20 times faster than the logical one.
But it by logic is a real fun! :)
Could someone explain the logic when at 1:28 (approx.) Simon decides than box 5 must be 567 and not 789?
I mean, I'm not sure if you have your question completely correct but it has to add up to 18.
@@Mitrasmit Doh! I forgot that, thank you.
Wow, this was the first puzzle I've done faster than Simon. And when I saw how long the video was I thought to myself, I might as well not even try it. I do think I got a little lucky early on though. I think I might have made an assumption that turned out to be true but didn't need to be.
Lol.
Funny one
Go to sleep, Simon.
[Yeah, count the sheep in your head -- or floating above your head -- going in circles -- which is what you had me doing with that "cursor" sitting on r6c6 @ 1:33:08
48:55 with a couple of dumb mistakes. I thought I ruined Box 6, forgetting I could enter it from Box 9 and then go right back out. I messed up the 4 in Box 1, forgetting I could move in a diaganol at the very beginning. And when checking my mistakes, I realized that I assumed the 6 in Box 4 was in Row 4 without considering it being in Row 6.
The line moving in a diaganol kept messing me up.
I also did this kinda fast on my phone, just going with my logic, not taking the time to prove it 100%, which ended up being the problem, haha.
I know in this puzzle the rules say sequential upon the line, but in other puzzles you’ve made this same deduction of the x-1+ x + x+2 = 3x and left it at that?
So how come you’ve never done x-2 + x + x+2 = 3x. So in other scenarios it could be 579.
As x -2 +x + x+2 still share the same properties and = 3x.
If the digits on either side of the central digit are the same distance away from it, they will always sum to the center digit x3.
Taking that to the extreme, 1,5,9 sum to 15 as 4,5,6 do.
We can go even further, 1 + 20 + 39 = 60 as does 19,20,21.
I only remember him doing it with three cell renbans (x-1, x, x+1), which also have to be sequential, just they can be in any order, so you don't know x is in the middle of the renban.
The rules say consecutive, not sequential. This resolved your dilemma.
I Solved it in 48:47min😁 I think it was the first time I was way faster than Simon😅
You know Simon's getting a bit fatigued when he starts providing explanations like "I'm going to put a pencil mark in there just because... I want to." (1:26:30)
The fly stole the show here.
To my shame, I ran out of patience after an hour of looking for a logical path, and found the solution by conjecture (trial and error). Must do better.
Simon did a heck of a lot better than I did at proving his solution logically, but did he have a small miss-step at 1:05:00? He seemed to overlook the possibility that the line from box 7 could have escaped to box 5 rather than box 4.
I have a feeling the three orange cell limit in box 5 means you run into difficulties if you try to go from box 7 to box 5, and still connect a path via every box. But it's not simple to prove.
My idea when I setted the puzzle was that you prove that the middle cell in box 4 is orange, so it's a 7, and then r6c2 is orange, consecutive to 7 and different to 6, so it's an 8. With this information, that 8 has to connect to something somewhere and the only available cell is the 7 in box 7.
If you want to see the simplest intended path I wrote a solution guide at this link: drive.google.com/file/d/1_dD8AX1iGVZxvnpWK5G6xlyGsl5GSrKP/view?usp=sharing ^^
(Un)fortunately, Simon forgetting about the possibility of diagonal lines doesn't result in error😅
Also (un)fortunately, Simon forgets about the 29 cell restriction or doesn't understand how powerful it is
I unpaused @30:54 and had to double check that I had actually unpaused. lol.
Very cool idea! Took me 55min
1:09:28 for me with a bit of help from the video to make sure I was going about it the right way.
Lol.
You cracked me up, Simon.
"Can we see any naughty 2×2s?"
(@1:33:07, btw)
And his cursor is sitting on one (r6c6).
What does Simon do?!
Alights on one up in row4.
Hilarious, Simon.
[I knew that center one had to be the 6 though]
Still funny, you alighting on another square when your cursor is sitting on a *definite* one.
Lol
👍👍😂😎☕️
Lol
That's where I was going wrong ("wrogn").
I kept getting a conflict.
That's *not* a "2x2."
Lol. 😂😲😲😂😎☕️☕️☕️
CORRECTED in another comment.
That's where I was going wrong (wrogn ) I realized when I kept getting a conflict on something else near the end lol .
Here we goooo
Us, looking back at this in a year: "There we went"
I solved this is 30 minutes with a secret trick I call "guessing"
Always prefer 'the orange' to be pronounced THEE orange and not th' orange with a glottal stop but thats just me! Hope this helps😁
SIMON, you inadvertently tricked me with your cursor sittin' on r6c6
[And I thought that was a "2×2"]
Lol.
I'll figure it out from here.
Thanks.
[I kept getting a conflict near the end]
I got tricked by Simon (and his cursor) sitting on a crossroads.
[And, boy, did I take the wrogn path, lol]
Fool by the "cursor" though, lol.
Good one, Simon (even though you didn't know it). 😂😎☕️☕️[
[[Oh, @1:33:08 when I cracked up mistakenly lol ]]
Here's a joke, Simon:
How do you "crack-up mistakenly?"
Like THAT up there ^^^
Lol 😂
wow, what a hard puzzle
"the way that can be wayed is not the constant way"
33:27 for me
39 minutes for me on this one
Solved it without any help from Simon!
34:30 kekeke he said doodoo
25:20 nothing, Simon! You're meant to do nothing so I can enjoy staring at you staring at the screen... for 8-9 hours straight if that's possible xD
69:38 for me. Marvelous puzzle!
37:57 for me.
I think it looks like an owl 🦉
As soon as you know the remaining boxes only have 3 orange cells each you can conclude that the central cell of the grid is orange, because anything else would leave a 2×2 of grey in the central box.
Edit: nevermind, I'm wrong.
where do I have to write to them for a birthday shoutout?
Send them an email via the email provided in the description
@@BryanLu0 Thanks! Don't know how I missed it!
Email is in the description
A little surprised Simon didn't see earlier the jump from box 9 to the 5 in box 6 and then back in to create the 7643 20 total, and then the 234 to make up the 9 total in box 8.
Woo!
25 minutes in: "This could be a short video."😅
Form the right thought-patterns? Better to go back to the simple point of asking the right questions.