Phistomefel Blows Our Minds

[@grimsqueaker5333]

The 3 inevitable things in life. Death, taxes and Maverick taking off as soon as Simon switches on his camera😂❤

[@chipsounder4633]

😂😂

[@LukeSumIpsePatremTe]

Do they both have fixed schedules?

[@joostvanrens]

​@@LukeSumIpsePatremTeyou mean death and trace, taxes and Maverick or death and Maverick?

[@sambartle3298]

I’ve always assumed Maverick is a pseudonym for any flight going over Simon’s home. Has he ever said it was the same pilot/plane 😮?

[@Daymickey]

😂

[@amoswittenbergsmusings]

Simon's closing remark about the feeling mathematicians experience when insights break through resonates with this old man. I have been working on a paper on the combinatorics of certain Lindenmayer systems and the associated directed graphs and adjacency matrices that exhibit stunning appearances of Fibonacci and Padovan sequences. For months I have struggled with the matter but lately things have been falling into place and revealed the underlying beautiful structures. At such moments the universe sings and you feel what the late great Douglas Adams meant when he put words in Fenchurch's mounth: "Well, that's alright then..." and Marvin in his dying moments is reconciled with his existence and shuts down in peace. Simon, you are a gift to humanity. Seriously. Und Herr Phistomefel, Sie sind der Goethe der Sudokuwelt.

[@malifex9922]

Solving this is one miracle. Imagine setting it! Simplifying, moving around the pairing boxes to the point that you need NO digits and only two small arrows to make a unique solution. Crazy stuff!

[@longwaytotipperary]

I’m amazed at how quickly Simon figures out how it works!!

[@leeannadimoulas5163]

wow. this is one of the hardest ones i've watched. it was tough for me to follow, which i dont usually find. Thank you for solving it! I found it kind of wild that you solved it using some discovered rule set, and barely even referenced the diagonal sums, instead looking at the parity and differences. The diagonal rule set up a whole set of secret rules you actually used to solve the puzzle. 🤯

[@ronjohnson6916]

Stares at grid. Yup. This is a nod along.

[@bobinvegas]

This is one of the puzzles where I could watch him on repeat 4 times and still not understand how this even begins. I swear these people are the next evolution of humans

[@PassionPopsicle]

I will also throw in some "Mhmm"s and "Sure"s and a few well placed chin-scratches

[@evergreentree8042]

​@@PassionPopsicle Gotta throw in a 'that's what I was thinking' for good measure.

[@insertsarcasm]

100% agree. I try to solve some of them but this one was like dude I'd have no clue where to even start so yep observing it is

[@KSilverlode]

This is one of the first puzzles for a while that I entirely gave up on trying myself and just watched as Simon performed feats of genius solving in response to Phsitomefel's genius setting. Loved it, even if I barely understood it!

[@faddy91]

Kind of funny to see Simon's hair after the jump cut. The phonnecalls must have been a more intractable problem than the sudoku

[@abecede4389]

Phone calls with LOTS of head scratching, judging by your hair right after the interruption.

[@runnerman8]

I noticed that straight away as well.

[@LorisLaboratory]

Ahh so much Phistomefel lately!! love it!

[@erl137]

absolutely love the combination of the sudden switch in Simon's hair and the initial remark of "disaster!"

[@TheEricthefruitbat]

I don't usually watch anything over 50 minutes, but Phistomefel is always tempting, and this puzzle appealed to my mathematical mind. Once again, Phistomefel pulls off something incredible.

[@77kaczka77]

Just repeating simple equation:: Phistomefel + Simon = delight for us

[@Twiztedgoalie]

Even watching this entire solve, I have no idea how any of the logic in this made any sense

[@inspiringsand123]

Rules: 06:33 Let's Get Cracking: 08:23 Simon's time: 1h9m35s Puzzle Solved: 1:17:58 What about this video's Top Tier Simarkisms?! Phistomefel: 10x (00:22, 01:26, 01:33, 03:04, 03:07, 03:09, 03:12, 08:28, 1:18:34, 1:19:12) The Secret: 5x (09:05, 09:17, 09:27, 09:27, 09:42) Maverick: 3x (01:54, 02:00, 09:57) Bobbins: 2x (15:54, 56:20) Knowledge Bomb: 1x (1:02:50) Three In the Corner: 1x (1:16:10) ​Scooby-Doo: 1x (23:41) And how about this video's Simarkisms?! Sorry: 22x (02:02, 17:13, 20:21, 29:05, 29:52, 30:22, 42:11, 46:37, 48:29, 50:34, 55:43, 55:43, 55:43, 56:22, 56:22, 56:43, 57:06, 1:05:39, 1:06:19, 1:08:15, 1:09:27, 1:16:27) Ah: 14x (11:44, 20:31, 21:16, 21:16, 21:16, 23:23, 26:53, 27:14, 34:09, 34:09, 43:17, 57:23, 59:58, 59:58, 1:00:27) Hang On: 9x (11:51, 12:00, 26:59, 26:59, 53:50, 54:06, 1:00:19, 1:00:25) In Fact: 7x (03:12, 07:09, 59:51, 1:06:04, 1:06:28, 1:08:24, 1:08:42) By Sudoku: 6x (48:18, 59:20, 1:12:25, 1:12:38, 1:12:56, 1:13:44) Weird: 6x (18:09, 18:09, 48:54, 1:01:27, 1:17:34, 1:18:40) Brilliant: 5x (03:22, 05:27, 32:06, 32:06, 1:18:05) Incredible: 5x (02:52, 03:37, 03:37, 04:36, 06:09) Beautiful: 4x (26:53, 26:57, 1:18:12, 1:18:12) Lovely: 3x (44:21, 1:01:57, 1:02:01) Gorgeous: 3x (03:52, 28:01, 44:49) Goodness: 2x (1:13:50, 1:17:56) Naked Single: 2x (1:16:25, 1:16:43) Extraordinary: 2x (1:18:05, 1:18:09) Magnificent: 2x (1:19:03, 1:19:06) Obviously: 2x (22:21, 23:00) What Does This Mean?: 2x (53:19, 1:10:01) Pencil Mark/mark: 2x (40:13, 1:11:56) Cake!: 2x (04:25, 05:14) Unique: 2x (00:31, 18:23) Good Grief: 1x (45:10) What on Earth: 1x (46:49) Useless: 1x (51:25) What a Puzzle: 1x (1:14:47) The Answer is: 1x (23:38) Missing Something: 1x (19:07) Naughty: 1x (1:04:58) Ridiculous: 1x (15:54) Discombobulating: 1x (52:14) Shouting: 1x (05:10) Progress: 1x (1:02:04) Wow: 1x (42:47) Most popular number(>9), digit and colour this video: Thirty Two (7 mentions) One (111 mentions) Yellow (18 mentions) Antithesis Battles: High (32) - Low (18) Even (48) - Odd (10) Higher (27) - Lower (9) Column (40) - Row (29) 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!

[@averygaron994]

And 55 minutes on the phone

[@charlesgaskell5899]

Seemed to be a lot of "sorry"s this time

[@mstmar]

The work done at 53:25 to find the contents of r7c4 apply equally to r6c4. this means that r6c4 is also from 5-8 and r6c3 is from 3-6. having done that, it would have been easy to find the yellow cell in col 3 at 1:00:42. this makes the logic at that time a lot simpler

[@raysouth1952]

OMG! I don’t normally watch these long videos but the rule set was so intriguing I was hooked. What a puzzle! What a solve!

[@andremouss2536]

25:30 A simpler (and more productive) way to show that the difference is two is to consider the two sums of the seven digits at the top. If the difference was four, the total difference would be 28 and has to be compensated with an inverse difference in the two pair of cells at the bottom, but that's clearly impossible. And even with a difference of two, that makes a difference of 14 for the two pairs of cells at the bottom, which can only be satisfied with 12 in one column and 89 in the other. With the circle which can only be 3 or greater, that gives us the location on these pairs of digits : 89 at the left and 12 at the right, and that at the top, the left colums holds digits two units lower than the right.

[@RichSmith77]

I did the way you describe, but I'm not sure I would say it's any simpler or more productive than Simon's way? I quite liked Simon's explanation.

[@andremouss2536]

@@RichSmith77 I just said that this method directly gives Simon's last deductions about Box seven.

[@dinane]

I needed a nudge to get past my brain insisting to try to find set answers here. But once Simon identified the even sum pairs I got excited and plowed ahead. Thanks for unlocking my brain to solve a fun puzzle!

[@anaayoung9142]

More brilliant than create is solve such puzzle! Many times I have no idea to what is going on, you sometimes feel the same, but you can always solve it. It is amazing!

[@troymcdougal9010]

I would love to see a Phistomefel review of Simon's brilliant solve! The great man always points out the differences between the logic he intended and the logic Simon follows which helps me understand both sets of logic even better. Great solve to a Great puzzle.

[@sumyrda2772]

I'm guessing the intended solve path starts with noticing that all the higher digits stay to the same side of the chain of border squares, that the side switch happens within the gray, and that the differences have to even out. Then you know all the differences and can work out the bottom arrow and then the other arrow and the bit about the yellow number ... that's all you need I think.

[@sumyrda2772]

So it's just like Simon did it, except without the parity detour.

[@davidrattner9]

Just phenomenal from the one and only Phistomefel!! Simon shows such respect and admiration while solving!!

[@jasonmetcalfe4695]

Still scratching my head on this one regarding the blatant trust of the digits in r5 6&7 based solely on a theoretical 7 in the circle in r4c6 while he was getting the jist of the puzzle

[@fountainsowayne]

I was confused by this too, but it turns out that those digits hold regardless of what goes in the circle. Like try placing a hypothetical 9 in there. Then below becomes 5 (by 4 difference), and the adjacent cell, by 1 difference, is either a 6 (in which case end of arrow is 3) or 4 (in which case end of arrow is 5). It has to do with the cells at the intersection of the two border square rows/columns.

[@erlandodk]

Brilliant solve by Simon as always. How on Earth do you set a puzzle like this? Phistomefel never ceases to amaze.

[@chocolateboy300]

I finished in 258 minutes. This was one of the toughest puzzles I have ever done. I spent about thirty minutes looking for anything and was so happy when I spotted that the row 6/7 and column 6/7 borders had to have a difference of 1 for each domino that crosses the box borders. After that it did not get easier. I wrongly assumed that column 3/4 border also had to be a difference of 1, because I miscounted. After reverting back to that point after an hour of wasted time, I saw that a difference of 2 was actually correct. As I tried to figure out the last borders differences, I noticed a pattern that every difference equaled 9. So, for the column 3/4 border, the difference was 2 on the upper side of the gray box and 7 on the lower side of the gray box. This allowed me to calculate the final border differences. They ended up being 5 for the left and 4 for the right as min/max shows that is the only possibility. The ending took a lot of work, but it turned out to be that r3&4c3 couldn't be a 16 pair due to geometry of the remaining digits. I feel exhausted after this one. That was brutal. Great Puzzle!

[@blobz-1]

“One does not simply walk into Mordor!”

[@CaptainSpock1701]

Phistomefel's genius is not just in setting sudokus, but in getting Simon to pencil"Mark" like Mark does... and then some!

[@brianmcadam443]

59:20 to solve, just barely in under the hour. Absolutely brilliant construction, but I might need a nap to let this cramp in my brain relax. (and really, the first 45 minutes was just trying to understand how the rows of different blocks interact. The actual *solving*, if you time it from the first "real" digit was barely over 10 minutes)

[@mahmoudabdelghany7112]

The masterpiece of all masterpieces.. damn it, it was tricky to deduce all the extrapolated rules, but equally rewarding..

[@TheFifthsWord]

Doesn't come up often but Simon's deduction @44:50 had been sitting there for so long while he tried to sort out the other arrow and I was just waiting for that 4/5 pair!

[@erlandodk]

I had that little area worked out before Simon! Mind you, that was the *only* part of the puzzle I had worked out and I would never in a million years have broken into this puzzle. But small victories I guess :-)

[@TehFilmFanatic]

I solved this when it came out and my god it was an exercise in building lemmas into a proof. I look forward to seeing if Simon ends up reaching the same pieces of logic I did.

[@roccov3614]

Braid, now that takes me back. Very artsy for it's time. I remember also being introduced to another amazing game about the same time. I don't remember what it was called but it was a quite enjoyable game in the first part and then you get bitten by a snake and everything changes. All sorts of new weird game mechanics and the game becomes 10 times better.

[@MatthewBouyack]

I usually just watch Simon solve the Phistomephel puzzles, but gave this one a shot and the title doesn't lie. My goodness is that brilliant! Got a bit stuck in the middle (on sudoku, of course!) but finally managed to finish it up with a time of 2:09:01.

[@andrewcook4873]

I got the main restriction here fairly quickly. The common difference means that the remainder that you can’t fit in a line has to go on the other side of the grey square. So the position of the grey square gives you the differences you must use on either side of it and the differences on either side of the square must add to 9. But then I got horribly bogged down because I’m not good at sudoku.

[@Paolo_De_Leva]

Again, *multi-versal* class sudoku. 15 minutes of standing ovation by all the members of the *Organization of the United Multiverse,* including the representant of the most hostile universe. 👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏 Brilliant solve by Simon (especially his impressively quick deduction @49:11). More importantly, excellent analysis of the breathtakingly beautiful *meta-logic* used by Phistomefel to build this beast 😏👍👍👍👍👍👍👍👍👍👍👍👍 This video deserves billions of viewers (in *each* universe of the United Multiverse, of course).

[@sumyrda2772]

Finally a Sudoku where I got on the right track before Simon did. Thank you Simon for this boost of my confidence 🤗, but also for the great explanations as always, and for making the mini German whispers green on instinct, before you even knew they were German whispers 👍

[@sumyrda2772]

That said, I'm a mathematician, so I might have had an unfair advantage here.

[@andremouss2536]

59:00 At this point, the 2x2 square with most power is the square R6C6-R7C7. In this square there must be a difference of 1 between all touching cells, and that may only be get if exactly one of the diagonals holds the same digit. But that means that this digit cannot be 1 9 4 5 (there cannot be two diagonals holding the same digit because then the difference woud break).

[@Paolo_De_Leva]

Right, I immediately focused my attention on that square, and the pair of identical digits it must contain, as I explained in my separate comment.

[@marcbennett9232]

braid is good but I havent played it in a long time. I remember some platforming but it was pretty mind bending too.

[@viviencollomb-clerc5526]

Incredible solve by Simond. Very nice.

[@thoughtbyte]

Brilliant puzzle, brilliant solve. As always!

[@MattYDdraig]

44:04 Ingeniously wicked and beautiful. A very satisfying puzzle in that there was so much to unpick and fathom, but once resolved it started to unravel smoothly. I was still sittong at 4 digits in after 35 minutes, before the floodgates finally opened in a delightful deluge.

[@steve470]

44:18 for me. That was some nifty logic to figure out, and it was pretty well telegraphed (albeit requiring playing with some numbers to get the feel for it). Of course, when I say it was pretty well telegraphed, I mean there are combinations that had to have a difference of 1 (with 19 pairs at the edge), then extending that logic a bit requires a difference of two (with 1289 at the edge), then you're prepared for the weirdness in rows 3 and 4. And, Simon being Simon, of course he spots the intermediate deduction before the easy ones. 😛

[@Paolo_De_Leva]

True 🤣

[@arnoudh6203]

Braid is great!

[@Robert_H.]

Let be the following a chain of boundary squares: a | b | c ------------ x | y | z Now applies: [1] a + y = b + x => a - x = b - y [2] b + z = c + y => b - y = c - z _______________ => a - x = b - y = c - z = const. = C => a = C + x, b = C + y, c = C + z *Rule 0:* Since a border square contains the diagonal sum S twice, a border square adds up to an even digit: 2*S. *Rule 1:* For a chain of border squares, the difference between the numbers of directly opposite cells is constant. *Rule 2:* For a chain of border squares, one side always contains larger numbers than the other. In the following, we refer to one side as hot (larger numbers) and the other as cold (smaller numbers). If the difference of a hot-cold number pair is a constant number, then the difference is either even or odd. The sum of a hot-cold number pair has the same parity as the difference. *Rule 3:* For a chain of border squares, the parity of the difference and sum of a hot-cold number pair is the same and fixed.

[@sergeilatushkin4979]

One of those rare cases when I think me and Simon could make a good team 😄Figuring out the difference approach (including the idea of 4x=5y, 2x=7y and so on) took me a couple of minutes, a minute later I've got the 9 2 8 1 square at the bottom... And then I got stuck for like an hour 😐Brilliant puzzle, as always. I've had a lot of fun twice (solving and watching) 💖 P.S. I suppose, the most vigilant watchers may deduce the amount of time Simon spent on those phone calls looking at the solving time (no spoilers provided)😉

[@CaptainSpock1701]

I've only start watching this solve (about a minute and a half in) and I can confidently say that I get the rules but have no clue about the beginning. This should be exciting.

[@RichSmith77]

I reached Simon's position at 59:00 into the video on my own, and since that's well over two thirds of the way through Simon's solve, I'm pretty pleased with that. I couldn't see how to progress though, so I ended up bifurcating here. I thought a 1 in r6c9 looked most forcing so tried it. (I did have 1 in box 5 restricted to r4-5c5. It couldn't go in the bottom row as 2 couldn't partner it across the boundary). One deduction after another flowed until 10 or 15 minutes later I had a completed grid. I now needed to rewind and see why the other option broke. It did, but only about twenty deductions down the line. Not too satisfying. I knew Simon would find a better solution. Even knowing the solution, I tried to find other weak spot from that position at 59:00, and couldn't find one. Eventually I watched Simon's solve. I'm consoled that the 2-3 deduction wasn't straightforward to see, and required a bit of a chain to rule out. But still, a simpler solution than anything I saw. I guess ruling yellow 1 or 2 into r5c3 first would have been the smoother path. How so little information in the starting grid results in a uniquely solvable puzzle is incredible.

[@GrumpyCthulhu]

There ist an easier solution, but Simon came close multiple times and always turned at the last moment. The 9 in column 4 ist ruled out off row 1-2 by sudoku, r3 by pencil mark, r7 because it breaks the 89 pair, and r5/6 because its partnering 7 can only be in r4 in box 4.

[@RichSmith77]

​@@GrumpyCthulhuAh. I see. Thank you. If I have time later, I might look to see if I can use that to finish from here. 👍

[@kolda89]

Wow 🙈 great solve… i wish there was more videos with setters going through the process… there was only one like that if i recall, right?:)

[@roccov3614]

I can't see how to start this. I'm going to have to watch the start of the video and hope Simon gives me some ideas.

[@CauchyIntegralFormula]

43m29s. I grasped the "mostly constant difference across a border" consequence rather quickly, but that didnt get me anywhere for a long, long time

[@maht0x]

got to grips quickly, then forgot the diagonal rules, never change Simon

[@sillvvasensei]

It's a bird! It's Maverick! It's Sudoku Simon!

[@AndersHaalandverby]

Quick question; please no spoilers: can border squares overlap?, and is it all bordersquares, including some that may overlap with the gray border squares?

[@AlphaetusPrime]

Yes and yes

[@nathanmays7926]

This puzzle feels extremely familiar.

[@FelipeRodrigues-vj1zb]

I actually understood very little of the video today, but it was still interesting 😂

[@LiquorStoreJon]

How does one even come up with such a rule set, then implement it into a puzzle??

[@efa666]

We don't dare fathom the depths of Phistomefel's brain.

[@leosmith848]

I think you work backwards from set properties of numbers.

[@aronlide]

Oh yes! BRAID!

[@omranshtiwi9324]

The fact that the 7 he threw in at 49:12 to test the theory turned out to be correct is mind boggling! to be honest I was anticipating a bobbins at any moment that would cause him to backtrack a lot or worse; abandon the theory because of the 5 and 6 over there! I know it didn't contribute to the solution a lot and he would've got it eventually, but pheww... IT WAS CORRECT!

[@RichSmith77]

He didn't keep his thrown in 7, though. He just kept the 35 pencilmark in r5c6 because he saw that that would be true whatever digit he had put in the circle. A few people in the comments thought Simon got lucky here. He didn't. He spotted a general pattern that meant combining the differences of 4 and ±1 resulted in r5c6 always being 3 or 5. That's all he used going forwards.

[@feldinho]

Gotta say this one went competently over my head…

[@efa666]

I'm hoping Simon gets inspired by the Braid soundtrack to do some guitar covers lol

[@mikew6644]

Whoa! This rules is wild… I’m going to at least give it a go Matt Parker Style!

[@VenomMorph]

number of times i screamed at screen that You can solve arrow and 4 digits inR89C34 is astonishing.... bet it would make the solve quicker. (btw not sure how it will develop, but I'm writing that comment after You said difference on R67 border must be 1 and totally ignored arrow there and focused on R34 border, around 33:45)

[@Arcessitor]

This for me personally is the hardest Sudoku I've ever had to solve. This broke me. 1:24:01 for me.

[@alunmurch4109]

Simon and Mark = Brian and Charles ❤❤❤

[@olliepeacock]

Must have been some stressful lift planning!! The way your hair suddenly became messy afterwards haha

[@skiesofarcadia4885]

I noticed that too and had a nice chuckle 😂

[@Paolo_De_Leva]

My strategy for solving included an useful aide-memoire in the rightmost four cells of *row 9.* After figuring out that the *arrow circle* in *box 3* is *r3c7* = *5, 6, 7, 8* or *9* I started testing its powerful interaction with the pair of identical digits contained in the peculiar *2x2 square* at the intersection between *rows 6-7* and columns *6-7.* In my aide memoire, I wrote the possible candidates for that pair: 🔹Cell *r9c6* = candidates available if *r3c7 = 5* 🔹Cell *r9c7* = candidates available if *r3c7 = 7* 🔹Cell *r9c8* = candidates available if *r3c7 = 8* 🔹Cell *r9c9* = candidates available if *r3c7 = 9* (I quickly figured out that r3c7 = 6 is impossible)

[@Paolo_De_Leva]

Unfortunately, I did not focus on *r5c6,* as Simon did @49:11. That was, in my opinion, Simon's most brilliant deduction and proved to be very useful later. My solve was slightly longer because I missed that deduction. In short, @49:11 Simon deduced in the blink of an eye that since *r5c6 = X - r4c6* (where X is the arrow circle) and *r4c6 = X - 4±1* then, by algebraic substitution, *r5c6 = 4±1* Simon repeatedly showed in the past that he can perform *algebraic substitutions* in his mind very quickly and reliably. In this case, he did not even bother to explain this fully. Maybe he thinks they are obvious for others as well... It was not obvious to me. Luckily, *Richard Smith's* comments below helped me to understand this *crucial deduction.*

[@craftyraf]

There was a lot of hair-scratching involved when comparing 46:05 to 46:10 🤣

[@puritan7473]

@craftyraf haha, someone needs to clip that with 1 second before and the words "right, disaster"

[@phballer99]

As a hobby video maker myself, I can't stress this enough. Make sure you stop recording for interruptions. Not because of editing, but because it allows you to quickly watch back the last minute or two of the first half and remember where you left off. 😁

[@rhysbart]

I would find it impossible to juggle all these plates even without an extended phone call interruption 😮

[@Gonzalo_Garcia_]

34:29 for me. Fantastic puzzle!! I'm pretty sure I had seen a similar idea before and that definitely help, but it still wasn't easy at all.

[@fulltimeslackerii8229]

Right off the bat, I think what this rule set tells us is that a 1 can never be diagonal to a 2 and an 8 can never be diagonal from a 9

[@jonathanbarz9898]

can someone explain at 36:00 why 5 times some difference has to equal four times some difference? i'm struggling to understand why the sum of their differences need to be the same.

[@jennyralgrievous6192]

I have absolutely no idea either, I'm just going on the assumption that it's some higher tier math I either didn't learn or didn't retain from secondary school 😬

[@kasperjuelsgaard848]

Both rows have to sum to 45, so if the difference between each of the five dominos on the right is 4, that makes a total difference between all five of 20. In other words, the row with the low numbers need the last four cells to add up to an additional 20 compared to the high digit row for them both to end up af 45. There is however only four dominos left so in order for that to happen, you would need a difference of 5 for each. Hope that makes sense. Its a bit hard to formulate ☺️

[@smylesg]

It's the same reason the offset difference of the other sets is 8 (eight pairs of 1 difference, one pair of 8) and 14 (seven pairs of 2 difference, two pairs of 7) to balance them.

[@skiesofarcadia4885]

I will try my best :) 1. It's important to keep in mind the "secret": Every row, column, and 3x3 box must contain the digits 1 through 9 once each (which together sum to 45). Therefore, the difference in value between one full row and another full row (i.e., 45 - 45) is 0, which means the total difference between row 3 and row 4 is also 0. 2. Within this puzzle, the border square constraint forces certain connected pairs of digits (such as Simon's five blue lines) to have the same difference in each pair. I won't re-explain here, because Simon demonstrates in the video, but in the case of Simon's blue lines in rows 3 to 4, the difference must be even. The smallest even difference is 2 (because 0 would mean a repeated digit in the column), and the greatest difference is 8, because 1 and 9 are the most extreme sudoku digits, with 4 and 6 lying in the middle. Therefore, the difference of each of the five blue lines must be 2, 4, 6, or 8. 3. Because the five blue lines will all have the exact same difference, their TOTAL difference will be the difference of each blue line multiplied by the number of blue lines, which can be simplified as 5x, where 5 is the number of lines and x is the difference of each line (which is one of the four even digits listed above). So, the total difference is either 10 (2 * 5), 20 (4 * 5), 30 (6 * 5), or 40 (8 * 5). 4. On the other side of rows 3 and 4 are Simon's four green lines. Their difference isn't necessarily even, but the difference of each green line is the same as each other green line AND the total difference of all four green lines is the same as the total difference of all five blue lines. The four green lines can be written as 4y, where 4 is the number of lines and y is the difference of each line. Therefore, 4y = 5x, or: 4y = 10, 20, 30, or 40. 5. Of those options, only 20 and 40 are divisible by 4 without any remainders or fractions, so they're the only two valid options, but 10 (40 ÷ 4) would mean the difference of each green line must be 10, but there's no way to get that high of a difference between two sudoku digits. Therefore, 5 (20 ÷ 4) must be the difference of each green line, and four the difference of each blue line.

[@khoshiumi]

How can i find Sudokus like this withouth having to rely on the link's provided by CTC? Somewhere i can see the dificulties he tends to provide and such

[@TurquoizeGoldscraper]

55:22 for me - Wow, that was something special.

[@rampantunease6517]

33:27 I solved it to this point then had to watch the video, Simon did the same path as I but here I got the four digits pencil marked in lower left, Maybe I am wrong 😅, because I stalled trying to work out rows 34 and columns 67 and humbled I am watching the video

[@rampantunease6517]

36:30 ah I was so close but couldn't see it and Simon explained it so well

[@aartbluestoke3352]

"what's the minimum clue set to specify a unique sudoku" is so much more vague than anyone ever thought ...

[@VacantPsalm]

Holy crap, I did it. Took like 2 hours, but I did it. HINT for those who might be intimidated: It requires VERY LITTLE busy work math. Once you figure out the trick, you don't need to do a massive amount of "find all the ways two numbers can add up to 13" over and over again. This is actually a good puzzle for those like me who are bad at that kind of math. (Is number dyslexic a thing?)

[@Situayo]

Dyscalculia.

[@xXFoiXx]

I can not for the life of me figure out where he sees the parity rule come from.

[@peaoui165]

I got confused with these rules. I wasn't sure if a border square could overlap another border square or contain any grey cells. Might be worth clarifying.

[@steve470]

They can and they can. "For all possible border squares except for the four entirely marked in grey..."

[@adamheywood113]

Sorry, I'm lost, how do you know where the border squares are? Do _all_ cells along the borders have to be part of border squares?

[@colej.banning2419]

Every 2×2 that crosses a border and isn't grey is a border square.

[@roccov3614]

I keep seeing patterns but not enough to solve it. For instance, I noticed early on that dominoes along an edge had to have the same difference and that difference was constrained by how long the run of dominoes is. I also noticed, while watching with Simon, that on one side of the grey squares you had even dominoes and on the other odd dominoes. The grey square marks the transition. Patterns but I think this one's too hard for me.

[@Ardalambdion]

Simon should do a Top Gun theme on the guitar honouring the Maverick.

[@smylesg]

Anyone else a little put off Simon didn't use green for the 5-difference line?

[@lucasmandato4910]

Am I dumb or he got lucky in the second arrow? Btw sorry to point out, love the vids ❤

[@dontmindme2844]

Absolutely...his second arrow was just the example if the number was 7. Its only limited to 23 or 56 if its 7! Some of the logic carries over. His 1 or 2 must appear in r5c3. That part is correct...but he would normally have to follow different logic strands to get a foothold down columns 6 and 7.

[@RichSmith77]

When he says, at 49:30, that "that should be consistent" I believe what he's saying is that the logic that gets to 3 or 5 can be generalised. It's a difference of 4, combined with a difference of ±1, to get a total difference that's always 3 or 5. He didn't articulate his thoughts too clearly, but I believe he knows at this point it's always 3 or 5, and then he verbally checks that a 6 in the circle would fit with this understanding also.

[@lucasmandato4910]

@@RichSmith77 ohhhhhh rewatched, i was dumb yeah yeah wasn't lucky...thanks for pointing out, i think i had lost its generalization while watching yesterday, thanks XD

[@paullackey8813]

I had so much anxiety over him leaving his example pen marks in until I later realized they were generally true.

[@OoHZa3376]

53:10 I swear I don't have a colour blind but I can't see the difference. 😂

[@Tahgtahv]

I am colorblind, but in a way that green is slightly more sensitive. And in fact the the vertical lines are greener, while the red and blue components of the colors are essentially the same.

[@imatrot]

Is it just me or is the arrow in columns 2 and 3 a lot easier to break into than the other one?

[@imatrot]

OK Simon saw it eventually

[@Raven-Creations]

Your deduction that the difference in C3/4 was 2 could have been much easier. There are seven dominoes, which means the high side is 7X greater than the low side, where X is the even difference. That 7X must be compensated by the two cells at the bottom of each column, which can have a maximum difference of (8+9)-(1+2)=14, therefore X must be 2, and the bottom of the high column is 12, and the bottom of the low column is 89. Because the arrow can't be 1 or 2, we know the 89 pair and the low side is in C3. "It's a lot of pencil-marking, I hope you'll forgive me" - Why do you never seek forgiveness when your refusal to pencil-mark makes a video twice as long as it needs to be? The breakthrough for me was asking whether R3C6 could be more than R3C7. I'd already got them down to 678 and 789, so the only possible way was if R3C7=7 and R3C6=8, however, that would make R5C7=2, then R4C5=2, and so R3C5=6. This puts 678 in R3C5-7, breaking the 678 in C9. Therefore, as expected, C7>C6. That was pretty tough, because it relied on having whittled down the options in R3C6/7 first, which was not trivial. Unlike when you worked out that R5C6/7 were 56, this did a lot for me thanks to my earlier eliminations. I wouldn't say it collapsed, but it got much easier after that. I did worry for how you would get on, because the latter stage did require a lot of observation and following up on deductions, neither of which are your forte.

[@thibauldnuyten2891]

I'm 26 minutes in but isn't this whole section about high and low numbers based on a speculative assumption? At some early point you assumed that "this side is always higher than this side by 1", which need not be the case at all? If you place the arrangement on the horizontal bars in columns 3 and 4 it becomes 2. Which by the way I'm really surprised you didn't do because it would have condensed the last 10 minutes into 10 seconds.

[@Sinisest1121]

10:10 Yes that was bird but his name is Maverick

[@RoderickEtheria]

This puzzle does look familiar. That said, I know it isn't the same puzzle.

[@sorellla]

I was about to write if this wasn't an older puzzle 😂

[@kevinsurget8571]

Simon: they must balance, 4x5 against 5x4, or 7x2 and 2x7. Me: 4+5=9, and 7+2=9, and 1+8=9, isn't it simple?

[@daniellemetz1433]

I don't understand why the rules say the 2 diagonals sum to the same when it means the two halves have the same sum... or maybe I just don't understand math's... but the rules feel incorrect 😢

[@RichSmith77]

It IS the two diagonals that have the same sum. Not sure where you were seeing two halves with the same sum, because that's not the case. E.g. the first 2x2 completed across a box boundary at 44:50 is, 9 2 8 1 The diagonals are 9+1 and 8+2, both summing to 10. (The halves would either be 9+2≠ 8+1, or 9+8≠1+2).

[@ilyrm89]

I didnt attempt the puzzle, these are out of my capabilities.. I could understand the whole difference logic quite well, but I would have never got the point that it needs to be 5 and 6 in row5, it seems too deep, which makes me to wonder if there was another route to progress.

[@mbarbh]

I love Simon's critical thinking skills. That said, I hope he goes back to 49:10 to see that he didn't earn c6r5 or c7r5, it was lucky that he chose the 7 in c7r3. I think he'd agree.

[@olivier2553]

If I did not make a mistake, for any possible value of c7r3, c6r5 would always be 3 or 5. Simon tested only 2 values and got the same result. So I was curious to check the other 3 possible values and you got it. But it was still a fluke.

[@japanada11]

The reasoning generalizes extremely easily though. If c7r3 is n, then c7r4 is n-4, so c6r4 is either n-3 or n-5, so c6r5 is either 3 or 5. I would guess Simon realized this even if he didn't work it out fully.

[@japanada11]

Note that Simon says "that should be consistent" at 49:30 - suggesting that he did recognize the pattern was going to hold for all values, and just checked one more option to confirm his guess. So while I agree that he didn't provide a full justification for his reasoning, it clearly wasn't a fluke either - he was confident it was going to work for all cases.

[@dontmindme2844]

@@japanada11 If he had really gone through the logic, it should have been very clear 6 didn't work because it puts either two 3's or a 6 in the same column with the 6. It appeared to me he tried two examples to get a general sense of how it works, but never proved it out. He really needed to test 9 and not just the lowest numbers with the fewest options down the arrow.

[@dontmindme2844]

And especially after he misses the obvious that the logic in r7c3 and r7c4 is exactly the same as r6c3 and r6c4 through the same mirror logic depending upon which row is higher.

[@chipsounder4633]

At 33 minutes, how can row8 column 3 be an 8? As the pairs along the 2 rows have to have a difference of 1.

[@brettduffin8412]

Is it a bird? Is it a plane?

[@gildasletallec3312]

I thing deduction around 1:00:00 might be wrong, beacause yellow can cross and is not 1;2: if you put r3 : 1;3;2;4, r5 : 9;1;5;7, r6:2;4;3;5 and r7 : 3;5;4;6 it works (for a while... before being broken by something else). Anyway 3-2 force a 3 in r6c3, then 5 in r6c4 and a 9 in r1c7, so there is no way to put a 9 in r3 and a 5in r4 in same column.

[@iodred]

I believe his deduction is correct. Idk what you mean by cross. If you put 5 and 3 in r5 and r6 c3 like you suggest, you dont have anywhere to put yellow in c3. The whole point was that yellow (being 1 or 2) cant be in box 1 due to the other 1 or 2 being r3c3, so for c3 to have yellow, it has to be in r5/r6 c3, and that can't work if the r5c6/c7 domino is 3-2 as proven by Simon.

[@gildasletallec3312]

@@iodred Ha. I hadn"t understood it well, maybe because of my average english. I understood both yellow were 1 and 2... Thank you for your good explainations !

[@angec9908]

Maverick is late today 🤣

[@Petrus74-yj4kv]

Simon's inability to notice his own pencil marks is far beyond ridiculous. One starts to worry his brain really has some defect...

[@RoderickEtheria]

1:01:24, I swear if Simon concludes that since those are nor 23 that they must be 56, I'm going to complain. 1:02:10, Okay, I'm complaining. At no point did he prove those were the only two options. He just leaped to that conclusion! He only had those down as possible options on an assumption of the digit in r3c7 being a specific number.

[@RuteNL]

No he found that r7c4 is 4 lower than the arrow circle, and r6c4 is 1 different from that. So the difference is 3 or 5 which needs to be in the tip of the arrow. If the diff is 3 then left is high in the vertical 2x2 squares so it goes with a 2, if the diff is 5 then the left is low and it goes with a 6 to its right. I feel like that was explained in the video Although the explanation was a bit sporadic

[@RoderickEtheria]

@@RuteNL He tested those numbers in case of 7. He did not test them for cases of 9, 8, 5, and I'm not convinced about 6.

[@pacman52280]

I was looking for a comment exactly like this one. He only did the actual math for if the number in the circle at that point was 7. He didn't do any other numbers, like you said, with the 5, 6, 8, or 9. I sincerely hope he gets it wrong and has to redo the puzzle starting at that moment.

[@TheFifthsWord]

@50:00 he shows that the 3/5 & 2/6 are consistent regardless of r3c7 because of what we know about the ratios. Ex if you put 8 in r3c7 then r4c7 had to be 4 away (so it's 4) and then r4c6 is 1 from that (3 or 5) and then 8 - (3 or 5) means r5c6 is 3/5 again

[@tylerhloewen]

@@RoderickEtheria He actually did test them for the other numbers, see starting at 49:30 he just did it verbally, and in his head without marking the numbers on the screen. So, the pencil-marks are earned, it just wasn't spelled out as cleanly for the viewer.