MPMSolution: Can you spin the table?

[@NasheedPog]

Participation points? Not getting a correct solution but giving it a go? *insert obligatory Parker square joke here*

[@volodyadykun6490]

Final puzzle: find Parker square but working

[@Vodboi]

They could be called parker points!

[@kfitch42]

The amount of point you get for giving it a go should be a square number

[@lsj8]

"Some people submitted a correct answer and stopped." -- because that's all that the answer submission page permits.

[@imdoingmypart6208]

You can refresh and do another one

[@lsj8]

@@imdoingmypart6208 ... which doesn't add any fields to the submission page. I'm not sure how that addresses the issue.

[@Khronosian]

To rephrase, "Some people completed the form, and did nothing else afterward", and this entire video was about the people who DIDN'T stop.

[@sabriath]

He is saying that if you submit a correct answer, you get the "points"...but if you show more, then you get featured.

[@srwapo]

As an engineer, I stopped once I got something that worked. Anything more is a waste of time!

[@matsuretro]

I had 1,3,5,7,2,4,6. My reasoning was simply, "Well, we can't have any two numbers next to each other be sequential, so let's skip over numbers."

[@kane2742]

I did the same thing. Thought of the reverse-order solution after I submitted this one.

[@Stephen-Fox]

Yeah, that's what I did as well, increase each person by 2 while increasing each envelope by 1 or vice versa. With 7 being prime it should work for increasing the people any number other than 1 or a multiple of 7, I think? The popular reverse order solution is equivalent to increasing by 6 each time.

[@brighamshamrell4524]

That's what I submitted as well. I arrived at that by the 8 queens approach.

[@AalbertTorsius]

I love how you're beaming with pride when you talk about Lucie's accomplishments.

[@Macieks300]

I hoped that we would also get the distribution of answers. I want to know which answer out of 133 was the least and most picked one. I assume the most picked one was the one with reverse order.

[@kyay10]

"The solution closest to Tau. I mean, at this point, it seems like a waste of time." Lol, it seems like u still hate Tau since that Numberphile video lol.

[@codeinnovator]

I used it as an inside joke. He noted it.

[@DavidLindes]

Aravind M: good work, Aravind! And based on my read of 7:14 in the video, I hereby speculate that he's actually coming around (or has even been fully convinced), and is just playing up the sourpuss for comedic effect. Either that, or he's actually feeling a bit of sourpuss over it, in which case I hope he will get over it. Pedagogy will be much improved for the eventual embracing of C/r as the circle constant, so... any promotion of tau is a good deed in my book. Thank you. :)

[@baksatibi]

15:01 Is that video related to a juggling cow sitting in an office chair?

[@lego312]

Something that made sense to me while looking at the solution is that: in fields with odd characteristic (Z_{2k + 1}), two generates the field, whereas it does not in fields with even characteristic (Z_{2k}) since 2 = 1 * 2 = (k + 1) * 2

[@codeinnovator]

One word, Honoured.

[@demarcheofedacity]

I should have sent my proof for not being able to do even numbers in. I used a nice bit of group theory for it. You can view your ordering on your table as an element of S_n and your rotation of the table as multiple actions of (1 ... n). If we can show that any even permutation has an n-cycle in its orbit under the group action of the subgroup generated by (1 ... n) then, as (1 ... n) is odd when n is even, any potential solution for n being even can be rotated to an even permutation and so can be rotated to a position where no locations match. To show that we have an n-cycle in the orbit requires checking two cases and then showing that these cases are sufficient. Unfortunately youtube comments arent the best for this. We will consider (1 ... n)(1i)(jk) with the cases j

[@element118_5]

12:55 For N-queens puzzle equivalence, I think we need the following transformations: 1. It doesn't have to be a torus, it just needs to be a cylinder. 2. You need the queen with only 1 diagonal (without the other diagonal)

[@adamsbja]

The submission page worked so well for thinking about the form of the solution that I thought you'd done it on purpose.

[@clockworkkirlia7475]

That's such a cool story about an extremely cool jacket! I also love the give-it-a-go attitude... I was always a bit of a square. ;)

[@Reddles37]

For the dice puzzle: 1) You can easily get a different die by just swapping two faces. There are 6! ways of drawing the faces, but you have to divide by 6 for rotating the `1` to a different face and divide by 4 for rotations around the axis, giving you 30 distinct dice. 2) It is not possible for only one side to match at a time. Suppose you put the dice down with the 1 facing up. Then each die can have a different number on the downward face, but that still leaves 3 numbers that must be on the side faces of both dice. If you just rotate them around the vertical axis then you can always get at least one of the side faces to match up, giving you two matches.

[@OscarCunningham]

Here's how I thought about the proof for even numbers. It's similar to Felix Broman's proof, but perhaps a bit easier to visualize. Proof: Suppose we have n people sat in such a way that we can only get one match no matter how we rotate the table. Now imagine drawing an arrow clockwise around the table from each seat to the person who should be sitting in it. Then each arrow must have a different length, because if two arrows had the same length then we could rotate the table by that amount to match both the people at the heads of the arrows. Since there are n arrows and n possible lengths, we must have exactly one arrow of each length from 0 to n-1. So the summed length of all the arrows is the n-1th triangular number, which is n(n-1)/2. But also, each seat has exactly one arrow pointing to it and one arrow pointing away from it. This means that the arrows must form into one or more cycles around the table. Hence the summed length of all the arrows must be divisible by n. We already showed that this length is n(n-1)/2, which when divided by n gives (n-1)/2. This is a whole number only when n is odd. Therefore it's impossible to find such an arrangement when n is even.

[@R2Cv1]

But you found n(n-1)/2 which is a whole number for even numbers. Dividing that by n is basically also dividing by 2 in the even case. That's like dividing a number by 2 enough times and then proving that it is not even.

[@OscarCunningham]

@@R2Cv1 I showed that if there is such an arrangement for n people then n(n-1)/2 divides by n. This happens only when n is odd.

[@niekgorissen9045]

I did a manual variant of what Felix Broman did. If you match each rotation needed to match (0, 1, 2, 3, 4, 5, 6 or -3, -2, -1, 0, 1, 2, 3 if you'd like) to an investor's number, there can never be two matches. To make sure numbers don't overlap (1 two forward and 2 one forward) I summed up the investor numbers + the rotation they needed to match (smallest in mod 7) and checked if they contained any duplicates to make sure the orientation exists. That way you can go from all 7! (or 6!) solutions to the existing solutions with max 1 match.

[@risfutile]

You know, I'm just here to let the algorithms know how cool this channel is.

[@frankharr9466]

Congrats on the launch!

[@rbradhill]

the community here consistently generates beautiful work. oh yeah, Matt does too sometimes.

[@PopeGoliath]

Beautiful squares.

[@AdrianWoodUK]

15:51 - Looking at Aric Parkinson's puzzle, here are my answers (showing my workings). 1: It is definitely possible to have a different die to my friend. In the simplest case, all I need to do is to pick two digits that my friend has draw on opposite faces, and draw them on adjacent faces. 2: It is impossible to arrange two different dies to only ever have one matching side at a time. Let's call my die "a", and their die "b", and let's assign the sides so that a1/b1 means the top face, a6/b6 means the face on the bottom, and a2/b2 to a5/b5 are the faces on the sides. We pick a digit to match, and spin the dice so that that's on top face of both, so a1=b1. We'll assume that in our arrangement, a6≠b6 (as otherwise that arrangement fails already), which must mean the digit on a6 must be somewhere on b2:b5, and vice versa. We'll spin our die on their vertical axis (leaving a1/b1/a6/b6 unchanged) so that a6=b2, and b6=a2, just to give us a good point of reference. So now, a1=b1, a6=b2, and b6=a2. The trick is, we now have 3 digits left, and the remaining faces for both dice (a3:a5 and b3:b5) are all on the sides. This means that, if we spin on the same vertical axis as before, we can definitely have all 3 remaining digits turn up on the same face on both dice at some point. This is the case for any arrangement that doesn't already have any opposing faces the same on both dice, which would already fail for that reason. Here's a followup question which I don't have an answer for yet; does the same hold true for dice other than d6s? It's definitely also impossible for a d4, but how about a d8, d10, d12 or d20? (Or a single d100, if you happen to have one of those monsters lying around?)

[@K-o-R]

Man, I never saw the second puzzle's video because it was on this channel. But now the third one is up on the normal SUM channel.

[@mojann1]

Yeeeees! This is the most necessary thing to ever be created during quarantine, turning it into qualitine.

[@skylark.kraken]

I just figured that what matters is offsetting the person from their seat a unique distance so I moved #1 1 space clockwise, #2 2 spaces clockwise, #n n spaces clockwise; and it all fits perfectly and was done and I think it works for all odd numbers but not even, and stopped after 5 minutes.

[@LeeSmith-cf1vo]

I always assumed there is some kind of limit on the length of video description, yours is pretty long!

[@GeneralYouri]

*Apologies in advance for the long post.* I was surprised that there wasn't any mention of factors, and how that ties into (a part of) the solutions, specifically those where you move a fixed n numbers every step. For the chosen group size of 7 this will allow n to be 2, 3, 4, 5, or 6. A value of 1 doesn't work as it simply returns the base sequence 1234567, but also *because it's a factor of 7*. A value of 7 doesn't work because it'd always repeat the same number: 1111111, again *because its a factor of 7*. This covers the only two factors of 7; values of 8 and up simply repeat the above pattern so can be discarded. Why is this relevant? Because when n is a factor of your group size, you are destined to collide with the starting point, similar to the animation you showed with the start that explained how even numbers don't work. For a move number of 2 you skip half the numbers, and for an even group size m you'll hit that group size after m/2 steps. Say m is 8, you'll create the sequence 1357 and then the next item would be 1 again, you cycle early and never reach 2468. For a move number of 3 the situation sort of flips; you create 14725836 which looks good initially, but index i here is always sorted the same as index i+4. So instead of hitting the same item twice, you hit the same *offset* twice. For prime numbers, every value of n works except for 1 and n themselves, as they are always the only two factors of a prime. For even numbers no value works, because 2 is a factor of every even number. For odd non-prime numbers there are also solutions, namely every value not a factor of the chosen n. For other numbers there are also solutions. For a group size of 9 you can use 2, 4, 5, 7, 8 as values of n. Note that 1, 3, and 9 don't work because they're factors, but 6 doesn't work either because there the offset changes by 3 per step instead, and again 3 is a factor. So there are two intertwined variables that can not be factors of your group size, and then it works. The amount of steps you move per item, and the difference in the offset with the item's correct seat per item. For a size of 7 and a solution of 1357246 the number of steps you move per item is 2, and the offset delta per item is 1: 1 is on its correct seat, 3 is 1 seat ahead, 5 is 2 seats ahead, etc. I said earlier that even numbers don't work because they have 2 as a factor. Well this is why - no matter what value you choose for n, one of these two variables is even, and the other is odd (because the correct seats are 1234567, always go up by 1). And I explained above that the solution doesn't work when either your step size n, OR the offset delta is a factor of the group size. And since one of these two variables is always even, it always has a factor of 2, and every even number has a factor of 2. Apologies for the long post, it was probably a bit confusing too, but I hope I got the number theory bits across! :)

[@lordofmorgul]

Also, loving these unscripted rambling videos. :D

[@jamesonhardy2126]

11:05 That's the bulk of what I was thinking about. I hadn't found the answer for even.

[@cheaterman49]

Whoa Lucy does space! Awesome! :-D stay safe both of you!

[@Adderkleet]

I know it's possible to have mirror d6's. And it really annoyed me when I realised it, because it meant I couldn't "nicely" line up all my dice.

[@ze_rubenator]

I wrote a C++ program to calculate it for n number of investors/salesmen, but there wasn't a way to submit it on the website and I couldn't be bothered to find another way. I still count it as a win for me.

[@jamesrayner9980]

You know, I'm just here to let the algorithms know how cool this channel is

[@ChrisWCorp]

I love the jacket!

[@NoNameAtAll2]

When are we getting "solution" of 3-way dice?

[@Krebzonide]

I GOT TOP 20 LETS GOOO!!!! I was kinda slow on the steam train because it was uploaded while I was asleep so I don't think I will be holding this position.

[@CheaterCodes]

Huh, I guess I should've wrote you a mail. You can definitely proof it for some even numbers: Let n be the number of seats at your table and i indicate the position of a person at the table. Now, assign the position i the number x * i, where x is the 'advancement'. This will work for any 1 < x < n/2, where x is coprime to n, as well as for n - x (other direction) The (incomplete) proof for this is: Only after incrementing by x n times will x * n be divisible by n again and create a cycle I worked this out in a bit more detail last time, but I didn't write it down, so this is quick and dirty. Some examples: 7 positions: advance by 2, 3 8 positions: advance by 3 12: 5 128: all primes < 64 except 2 210 (2*3*5*7 factorial): 11, 13, 17, ... (all primes bigger than 7 but less than 105) I think this way you should be able to prove for almost any number that it's possible, maybe even all positive integers.

[@CheaterCodes]

If you find a mistake, let me know. I *think* this will work.

[@huawafabe]

Weird question: can you do multi camera videos? :D I wanna watch you on my OLED and I'm scared of burn-in when the video is longer than 10 minutes haha

[@croco049]

I feel you should also give extra points for the quality of posted solutions. Extra points for using a emoji LUT in an Excel, extra points for making an animation to solve the problem, extra points for making an entire dedicated video with 3b1b-quality animations …

[@konkos5473]

on the prize for the people who gave it a go but didn't quite get it...Could you please call it the "Parker Square Award"?

[@Macieks300]

The link below "This is the gif which Jonas Lekevicius made" is not available

[@JonasLekevicius]

I've also tweeted it: twitter.com/lekevicius/status/1242901810072993793

[@Macieks300]

@@JonasLekevicius thanks

[@becauseimafan]

@@JonasLekevicius Thank you! It looks excellent!

[@jamesarthur67]

I knew it had something to do with group theory!!!! I sent in a document with me resenting about all its links with group theory XD

[@jamesrobinson6330]

You really need to learn how to use Manim by 3B1B, its great!

[@SmileyMPV]

What's up with the with the jacket?

[@codeinnovator]

the Dropbox link is broken.

[@bosstoober8782]

What happens if you submit an answer, and then realise you were wrong? Are you allowed to try again?

[@agmessier]

I'd love to be able to see how many correct answers have been submitted already for the current puzzle. It would be good motivation to try to be among the first 1000 if we didn't see the video as soon as it was released.

[@Tfin]

Just assume no one has gotten it right.

[@RedPandaSebas]

That is EXEL ent

[@jiaming5269]

Should title MPMP Solutions hahah MPMP is branded into my head alr

[@GoranNewsum]

So half points for a Parker Approach? Gotcha!

[@Grim712]

What happened to the coin that had the same probability to land on it's edge as on it's faces?

[@severnsevern580]

any chance i can check my submission and score?

[@keeperofthegood]

LMAO Matt, the cool part is to see in you in your smile and eyes, when you talk about your wife, how cool you think she and her job is.

[@k4it4n]

Do you have the time for how long it took for someone to get the answer, I thought I was pretty quick, and got the answer submitted second. Just thought it would be interesting to see how quickly people could solve the problem.

[@zacharybarbanell1064]

How do you want us to send in general solutions and such?

[@wolfbd5950]

Re. the jacket, would that happen to be the satellite which originally only went partway to space on the Ariane 5?

[@edwardlane1255]

no I think that's a different satellite

[@reddcube]

What happened to the 3sided coin?

[@francoistrempe]

Maybe I missed something, but how do you submit your proof?

[@aj_they]

I'm equal 1000th! One question I probably won't get the answer to, but if I put in a different display name with the same email, will my points be under my old display name "Avery 3" or the new one, "Avery Kae"? Or will they be under both and I won't get points very well.

[@zoegriffiths1308]

Don't worry if you entered a different display name for puzzle 2 - we identify you by your email address and will keep your display name as the first one you entered (or the one we emailed you) for the rest of the competition, regardless of what you enter.

[@aj_they]

@@zoegriffiths1308 Ahh that's a shame, I wanted the new one! I guess "Avery 3" is a good name as it is :D

[@Vodboi]

What's up with th... with the jacket? :^)

[@AndrewMacMe]

Tord Skiftestad His partner was involved in the ESA and NASA Solar Orbiter project.

[@dagordon1]

He mentions it near the end

[@becauseimafan]

Yeah he mentions it in the last minute of the video :)

[@theignorantphilosopher4855]

I'm gonna get so many participation trophys.

[@Ellyerre]

Nice Parker jacket.

[@diegopuppin]

The link to Felix's proof is broken, unfortunately

[@erwinjohannarndt4166]

Matt's PMS... Good name (?

[@JoeKoenen]

Half Points for a Parker Square!

[@croletan5424]

That jacket is lovely.

[@Nors2Ka]

You misspelled first name "Jonas". It's more like Yo-nas, the J in Lithuanian is soft. And also the "c" at the end should be "č", but I guess that's how you got the e-mail or whatever. I don't blame you though, we have a very unique way of pronouncing and spelling things :)

[@YellowBunny]

Here's an approximate copy of my comment with my solution from the original video: The problem can be rephrased in the following way: Are there any 7x7 permutation matrices that that have exactly one 1 on each main pandiagonal? There are in fact 19 inherently different ones that can come in 7 different rotations each which leads to a total of 133 solutions. I wrote a program in C that finds all the matrices for arbitrary dimensions and prints only one of the n rotations. For n=1,2,... there are 1,0,1,0,3,0,19,0,... or 1,0,3,0,15,0,133,0,... solutions depending on how you want to count them. So, it looks like there is such a matrix for odd n but for even n you can always get two matches by rotating. Anyway, here's my code: pastebin.com/cMnEm72m And here are the 7x7 matrices (rotated so the first person sits in front of the correct envelope): 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0

[@MarceldeJong]

I don't understand half of these solutions. But then again I also didn't fully understand the maths problem.

[@minijimi]

Easy problem to solve in a complex manner.

[@klikkolee]

The speed bonus introduces a huge bias towards schedules which align with your uploads. The spin-the-table puzzle could be solved by trial-and-error in about 10 minutes. Meanwhile, many people won't get the chance to see the video for almost 24 hours -- and that's assuming every day has an opening, which isn't always the case. For the first 48 hours *at least*, any speed bonus needs to be a constant value, and it should only decrease in multiples of 24 hours. Plus, lots of people will be interested in participating, but don't want to do so at the earliest possible moment that their schedule allows. A speed bonus definitely creates a "filthy casuals" divide. I wasn't even aware that the steam train puzzle existed until just now. Honestly, if the speed bonus sticks around, participating in this competition would require unreasonable lifestyle changes for me, and I am very doubtful that I'm alone in this regard.

[@lordofmorgul]

Why does Matt keep saying "giuplicates"?

[@jeepien]

For some unknown reason, Brits often pronounce a D as if it were a J.

[@lordofmorgul]

@@jeepien But Matt is Australian, afaik. :)

[@PaulPaulPaulson]

Matt's Mad Math Mayhem

[@EnergiaRocket]

odd even

[@unvergebeneid]

Paddy's code seems much nicer tbh.

[@sammys9361]

woo! I placed 3rd! I'm sammy 1 :)

[@flikkie72]

Better than #hometasking, this is!

[@stephenhousman6975]

TBH I found the 1,3,5,7,2,4,6 solution before the reverse order.

[@canislupus7722]

Drawing stars is fun. ;-)

[@schweinmachtbree1013]

18:10

[@ClydeHobart]

Even it's the case that everybody who submitted these extra bits also happens to use gender neutral pronouns, I wanted to say that it made me happy to hear him just using "they" in situations where many would just assume gender based on first names.

[@robo0428]

woah what's with the jacket

[@judahzf]

The PowerPoint is 4:3 😫

[@merlinmagnus873]

A good solution on the D20 problem: www2. oberlin.edu/math/faculty/bosch/nbd.pdf It's what I use when making them.

[@dcsignal5241]

Imagine having a Race where some People can't start unless they get up at 4am. Competitions suck.

[@officialEricBG]

hey what's up with the jacket?

[@MikeBucceroni]

didn't pay attention to the whole video, did you?

[@timothytraver5918]

Maths... Solutions??

[@chrisr7809]

Like for the cool jacket

[@simyann8563]

Typical anti-Australian sentiment from an Englishman, uploading videos in the middle of the night here in Australia so we can never be in the top 1000.

[@simyann8563]

What kind of accent is that for an Englishman anyway? Sounds like someone from Perth.

[@davidmoore1253]

@@simyann8563 He is Australian.

[@kstringer24]

1:55 "jiff" - i will correct this every time.

[@SoleaGalilei]

Haven't you got anything better to do? (Well, maybe at the moment you don't, to be fair...)

[@Tfin]

"Heave." Each letter's pronunciation varies by language spoken, and the creators forfeited the right to define pronunciation by their long silence.

[@kstringer24]

@@Tfin It only needed to be stated once to be defined! The earth is NOT flat, and I refuse to be silenced by the lazy and the mediocre!

[@Tfin]

@@kstringer24 Stated once, IN A TIMELY MANNER. They took too long.

[@gildedbear5355]

It's probably a good idea to re-establish the puzzle that is being solved so people don't have to go back and watch the previous video.

[@muxunmuxunov8409]

First comment is mine. Eee boyyyy