Zach Star yass queen

Here is a reason why the Euler Brick problem would not be solved. a2+b2=x2 a2+c2=y2 b2+c2=z2 a2+b2+c2=d2 What they didn’t say is that, through manipulation, 2d2=x2+y2+z2 and the thing is that square root of 2 is irrational, and so it would be hard to work stuff like that.

Other way to think about that is, what is the maximum amount of clues you can give and still have more than one solution. Add one to that and you will have a minimal sudoku I guess.

@@datguiser so sqrt(x^2+y^2+z^2) also has to be irrational (a multiple of sqrt(2)?) For this to work. I may be wrong, but it seems to me that x^2+y^2+z^2 is a non-perfect-square power of two.

Zach Star i actually did that 41 puzzle you showed on screen at 10:28 and got more than one solution but i don’t know if that was the point of it or if it was supposed to be minimal

me:o

me: :O

Me: :0

You can easily take this to infinity though... If the number line ranges from 0 to 1, you just put point number n at (n-1)/n + epsilon. Of course, you end up crowding points in very close near the right side, but nobody ever said they had to be equally spaced. I may be missing a constraint of the problem.

@@notnotandrew nice method but if you try it you'll see that it does not work for n>3

Anushka Uniyal bro 3am in my city and i have a lot of homework to do, and I’m watching this, totally feel you bro

@@FranciscoGonzalez-hz2bn Bueno Diaz ! "It's a her" Anushka is a lady. Indian

That's 5am for me and no, I'm not a morning person.

Literally me

Anushka Uniyal I am literally doing the exact same thing rn

No

It’s been a week, how are you feeling?

I’ve been bumping into things and I’m not sure if I’m in the right house

@@billywhizz09 you have my permission to blink again, I am sorry

omg thank you ahh that feels so good my eyes feel refreshed again now

lemme introduce you to a thing called piracy

Thats what I thought 😂😂

That's so true ahaha

Flow free is for the weak. I play flow free hexa

Mirian Raamat same but I finished all of them sadly

Happened to me aswell

How does it work?

@@ladripper47874 kzfaq.info/get/bejne/g8CGeK2n16fHhaM.html

Ditto

I didn't blink but I looked to his fingers and feel like I just missed something by looking there and not the whole hand with the card

I was looking for this comment lmao

That's the joke

I read this as he did it

He memorized them on paper.

weedbong_ same man. I knew someone would bring it up in the comments.

Anti-Science is on the Rise. Uneducation causes Muffled Logic to be be more and more accepted, so casual B.S. is getting more and more popular. People embarass themselves all the time now by claming NASA is faking the Sun, the moon is a hologram, the Earth is flat, Aura and Chakra are kinda Science, so trust me bro, i know we are all immortal - oh, and one last thing: Koalas are Fake; they are ALL CGI. All.

Lmfao

😂😂😂

Mhhhhhh👀

D is invetween

@@fredriklarsson1707 the line for d wouldn't touch the line for a though

​@@snifferrr Im sure that with the distance of solution, that the circle is a part of the line marked as a begining and endpoint to clearify the distance

@@snifferrr The lines for the Ds and As would touch at the point D on the perimeter.

We share a lot of courses with medical engineers, plenty of women there ^^

2 years into my electrical engineering career, and can say that this is false. My Configuration Manager is a woman, and I still visit my mom from time to time.

I became depressed 2 months into electrical engineering

Air Crash I'm CS, we have women 😎👍

I am a physics major and we have more women than men in my country. Too many are also bad.

Dunno who that is, but I’m still laughing cause I can only imagine Parker failed to make the first Euler brick.

@@Solrex_the_Sun_King That's Matt Parker, channel name "standupmaths." Also makes frequent appearances on Brady Haran's Numberphile channel. Look up, "Parker Square" for the reason for the main comment in this thread. PS: I forget whether the Parker Square video is on standupmaths or on Numberphile. Fred

This comment made my day😂😂

He will never live that down

Even on other channels, there's still references to the Parker Square. 🤣

No replies

So I replied :D

@@I_killed_that_beard_guy No replies to your reply

@@I_killed_that_beard_guy So I replied :D

@@commentor5479 :D

You can just do two pythagorean theorems. It's pretty cool when you learn that the pythagorean theorem extends to 3 dimensions like that.

But you always get pi

@@danielyuan9862 So I assume it extends to n dimensions.

@@ligafftheindifferent3495 4 dimensional pythagoras’ gets tricky, but it is theoretically possible - just don’t ever try it

@@gf1006 it’s really not though? like you can generalize the pythagorean theorem to any number of dimensions by adding another +x² term (like a² + b² + c² + d² + e² + … = h²) . that’s actually how you figure out the length of N-dimensional vectors

When you got 462 likes but no replies :(

Not Your Average Nothing but then the brick can’t exist realistically so...

FpS Blitzzz It forms a brick black hole.

Ur so big brained

Or the Infinity Euler Brick

El Emeno Pea infinity isn't an integer

What's it changing to?

@Canol Onar ( ͡° ͜ʖ ͡°)

How much prep did you undertake in making this decision? Might it have been a significant amount?

Perfect euler brick? A=440000000 B=1170000000 C=2400000000 D=2706011826 X=1250000000 Y=2440000000 Z=2670000000

@@kaelanmick3065 No: D*D=7322500002451854000 A*A+B*B+C*C=7322500000000000000

Yeah that and the fact that if you shuffle a deck then you likely have come across a combination that has never been shuffled before, very mind blowing to me. I learned that a while ago though and guess I was just focused on things I learned recently.

@@zachstar That's true in theory but probably not in practice-humans are really bad at shuffling cards

@@abijo5052 As long as you use a proper method of shuffling (spreading all the cards, moving them around, and putting them on a stack is the simplest one), yes, yes it's true. After all, a proper method of shuffling means you (almost) randomly take one of the 52 factorial ways of ordering, and 52 factorial is about 80658175200000000000000000000000000000000000000000000000000000000000. Which, for the record, is about 20000000000000000000000000000000000000000000000000 times more than the amount of seconds that have passed since the big bang (which is about 4000000000000000000 seconds).

@@abijo5052 Huh, I think YT deleted my comment for spam due to the amount of zeros I put into it. EDIT: Oh, it didn't, YT was just messing with me and not showing the comment. Anyway, you have two comments now. Anyway, yes, if you shuffle your cards PROPERLY (that means any method where there's significant variety in how the cards end up being ordered simply through the movements you make, the easiest way to ensure this is putting all the cards on a table, spreading them, shuffling them around, and then putting them back on a stack), it has a very high chance to be a unique method. 52 factorial is almost 10^68, which is about a quintillion (a billion billion) times more than the number of atoms that together make up earth. Like, you can mess up your proper shuffling badly enough to lose 20 factorial (meaning 20 cards are guaranteed to be in an order you have seen before) and you still have (to use the previous comparison) more options to order the cards than there were seconds since the big bang. The amount of seconds since the big bang MULTIPLIED BY the amount of seconds since the big bang options, to be precise. Well, and then again some ten times as much, but we don't count such little differences.

There is more chance of the atoms lining up and your hand passing straight through a table than shuffling a deck and getting a combination that someone else has already shuffled

I did the same xD I was thinking it was similar to another puzzle I've seen and just went with it

My A-A traced the circle lol

I thought about connecting on top of the circle’s circumference

I decided to see 13 cards in a suit, which is for 12 moons in a year with a spare for when there is 13 full moons. I like it can be used as a calendar. That it is not supposed to be is just more ingenious.

and where do you think the tarot suits come from? i can accept that 13! averages out to seven days per 52 is possibly somewhat coincidence, but the fact of the matter is that celestial clockwork was a driving or inspiring factor throughout all early cultural history until electrification.

@@TheJacklikesvideos the suits came from the Mamluk swords, coins, myriad(cup shape), and polo sticks, representing military, mercantile, spiritual?, and sporting. But the suits originated from chinese cards where the suits were: Cash (coins) -> coins/diamonds Strings of cash -> polo sticks/clubs Myriads of strings of cash (10 strings each) -> cups/hearts Tens of Myriads of strings of cash. -> swords/spades And the chinese cards were 1-9 in each of 3 suits (27 cards) possibly with 12 or 13 in the 4th suit (39-40 cards) if there was a 4th suit. The 1-9 in 3 suits continued to Mahjongg. It is interesting to look at this and realize how much changed based on there being only symbols on the early cards. As the cards moved westward, the images got reinterpreted, then redrawn, just like the game of telephone.

Another fun fact is you can shuffle any 52 deck of cards and you will most likely be holding a deck that nobody in human history has ever held.

​@@BiggerBearnot probably you really will be holding a shuffled deck that has never been before it's the 52! factorial

And the fourth one?

@@callisto5097 Genghis Khan probably

Pretty sure Genghis Khan isn't on a regular bicycle deck

francisco jimenez de cisneros, because nobody expects the spanish inquisition!

Rebecca Alderstorm bill

Russell Teapot look into group theory

Not all rules and conditions of this puzzle were intuitively given by the thumbnail. Namely, by "line" we intuitively assume "straight line" when the solution is merely about nonintersecting _paths,_ in which case there are actually multiple possible solutions. Mine for example: - Start drawing paths from D,A,C on the lower left edge, each of them circling clockwise around point B near the center. Paths D and C will connect to their endpoints while path A must emerge between them (and continues onwards). - Now draw paths with A and B as a pair, each circling counterclockwise out of the center. Path B connects to B on the edge, with path A on the correct side of path B to continue onwards and connect to point A on the top edge.

Hi. Know Sci Man Dan? The funny education-youtuber?

Anti-Science is on the Rise. Uneducation causes Muffled Logic to be be more and more accepted, so casual B.S. is getting more and more popular.

Lol

And that’s *if* one exists, so it may turn out to be not surprising.

Oh what???

The design on the 8 Card makes an 8

Nope

Not at all. That line pass through at least another point, and that's not allowed.

AAAAAAAAAAAAAAAAA

Make that, "playing contract bridge," and it's even better. Fred

But he's the joker.

Good one!

@@jayfredrickson8632 obama is the joker

It appears twice!

exactly what i thought :D

@@31.vaishanavikurup20 so? this does not affect the answer the commenter has given

@@kyyay2283 Did they remove their comment about being unable to leave the circle? If that's what was removed, then yeah it doesn't affect what I have said. By me connecting point B to B and D to D in a straight line, there is a small gap between said straight lines which means A can be joined to A and C can be joined to C by long curved lines.

@@Serpinstrix yeah, it was that

Surely someone must have thought about generalising the problem to further dimensions. But since the existence of a peefect euler brick is still unsolved, I doubt one can prove the existence of a 4 dimensional brick

Each of the ‘faces’ of a 4D Euler brick would be a 3D Euler brick, so my instinct is you’d have to prove the 3D case first. Unless you define a 4D Euler brick in some other way... Unsure what other definition would be reasonable for a 4D case.

Isn't that Fermat's last theorem?

nabiddy badiddy no, because whatever dimension you go in, you will only need squares and square roots to calculate lengths

@@toniokettner4821 ah yes of course you're right

good for u 👏

Ok boomer

want a medal?

It's actually very easy if you just connect "a to a" last

@MajorPrep: Thanks for the 💕, and for the interesting stuff in the vid. That problem of placement of points on a line in that way, is one of those "who'd-a-thunk-it" results. 17? Maybe this'll get you promoted to Lt. Col. Prep! Fred

7 years ago is recent in some contexts

@@drdca8263 True, it really depends on the reference point. According to Wikipedia the first ever Sudoku was printed in 1979 (however it was first popularized in 1984), so this style of puzzle is now 40 to 41 years old. Considering this, I would not claim 7 years (or more, not sure when this fact was truly found) to be "recent" xD

MoD366 Oh! I didn’t realize that sudoku was that young. Good point!

@@drdca8263 I also expected it to be older. Good thing I checked first xD

Yeah he thought it's gonna capture attention but it actually repel But I just skipped that, now everything is fine

Actually, it was quite interesting. Also, it wasn't numerology, but is, in fact, a historical part of how the deck of cards was first thought of and created.

@@roylavecchia1436 Agreed, the card thing was some interesting history behind I assume the number of cards in a deck, the number of faces, etc.

I solved it with a different solution too

No wonder I was confused (totally not because I'm dumb)

To a topologist, they might as well be straight lines, since the solution will be homeomorphic to one where the dtos are connected by straight line segments. Maybe they are straight lines through a distorted plane.

@@isaacwebb7918 so, the earth is flat, but the universe is curved. which makes the line straight even though it looks bent

It is straight if you are on a sphere. Three transform is rotating a sphere so 2d projection realigned the points

Yeah, and the odd edge must be larger than 2.5 x 10^13. Maybe not, for me.

Time for quantum computing!

is that really true

I'm not sure if I'm correct but 1 x 2 is 2 So 2 is the sum of 1 prime and 1 non-prime rather than 2 primes.

@@abdi165 thats a product. sum is when you add

@Parker Shaw A sum of two odds give an even number Despite most of the prime numbers are odd. 2 is the only even number that is prime. Hence the sum of any 2 primes doesn't really work as 2+3 or 2 + any prime number other than 2 is odd.

Just for better explanation of the problem, any even number CAN be expressed with the sum of 2 primes. There is no condition that says an even has to be expressed with a specific prime. This problem is called Goldbach Conjecture and you can find sources about it

I did the same - I'm guessing a frame was removed.

Haha thanks man! Not stopping anytime soon

If youtube still exists then

And that’s exactly why I’m replying

Brady Walker Do you feel smart now?

those are lines, welcome to topology.

@@boredommm23 if by topology you mean IQ you are correct

Curved lines:

exactly my thoughts

Hi. Know Sci Man Dan? The funny education-youtuber?

You're looking at a projection of a wiggly surface maybe a sphere even, so that is why transform works. Sphere projection rotates

Lines are a set of points that go to one end or another, it doesn't have to be straight.

@@dazzleneal true and false, depending on where you live

I did the same lmfao

it'd actually be really difficult, if not outright impossible to even write a functional program, given the limitations of how computers store numbers. From what I understand, the only theoretically valid numbers are so large that in order to properly calculate it, you would need a very large bit integer, as memory addresses (where numbers are stored in a computer) can only go so high before it has to resort to "compacting" the numbers, which is inviable if you need exact numbers rather than rounded or numbers stored as equations, both unable to give exact, solid numbers a Euler cube needs. To explain, most computers are able to store, say, 1^1000 + 1 just fine, but having that same number in numerical form is nearly impossible, as even a 512-bit processor would overflow (the maximum storage for memory addresses are 2^x - 1, where x is the number of bits the processor uses). Since processors become exponentially harder to work with the higher bit they are, and any use for such high numbers exponentially decrease, its nearly unheard of for software to go past 128-bit. (for reference, the original NES released in 1986 used 8-bit software, and modern windows devices use 64-bit). in addition, all of this is assuming that the memory address storage is the limiting factor, which in reality is extremely unlikely to be the case past 64-bit for reasons that I can't explain without delving into a bunch of hardware and software-related content that even I'm not fully versed on.

congratulations

lulukenn _ no

@@theyumblat5420 sorry?

He said it should be considered 1-dimensional

to be minimal, it requires that removing one number from that 80 will yield a sudoku puzzle with more than 1 solution and that is impossible

bronchosaurus211 mabye it isn’t the problem is unsolved haha

@@ryannoonan5518 It's easy to see why 80 is impossible. Removing 1 from 80 means 79 filled squares and 2 empty squares. To have two solutions, the two numbers used to complete the two empty squares must be different and interchangeable. But that cannot happen because the number that belongs to each square must be the 'missing number' either across or down (e.g. if one of the empty squares belongs to a row that contains all numbers from 1-9 except for 7, then 7 MUST be the missing number that completes that particular square and hence that 7 cannot be interchangeable with another number like 3 or 5 or anything else). Hence with 79 filled, there is only one solution. But with only 50 filled it's harder to tell since there are so many variables that complicate the problem. That's why it's still kinda unsolved

@@luvsYuri This seems really easy to do with 77 then. Create a board where the four open spots are a 2x2 area and the missing numbers are two 1's and two 2's. Make the two 1's in opposite corners and the two 2's in opposite corners. Then make two of the numbers in one square and two of the numbers in another square. Now both the rows, columns, and squares have a 1 and a 2, but you can switch the 1's and 2's and it will still be fine.

bronchosaurus211 oh yeah that’s cool nice