Veritasium - 2021-07-30
The Collatz Conjecture is the simplest math problem no one can solve — it is easy enough for almost anyone to understand but notoriously difficult to solve. This video is sponsored by Brilliant. The first 200 people to sign up via https://brilliant.org/veritasium get 20% off a yearly subscription. Special thanks to Prof. Alex Kontorovich for introducing us to this topic, filming the interview, and consulting on the script and earlier drafts of this video. ▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀ References: Lagarias, J. C. (2006). The 3x+ 1 problem: An annotated bibliography, II (2000-2009). arXiv preprint math/0608208. — https://ve42.co/Lagarias2006 Lagarias, J. C. (2003). The 3x+ 1 problem: An annotated bibliography (1963–1999). The ultimate challenge: the 3x, 1, 267-341. — https://ve42.co/Lagarias2003 Tao, T (2020). The Notorious Collatz Conjecture — https://ve42.co/Tao2020 A. Kontorovich and Y. Sinai, Structure Theorem for (d,g,h)-Maps, Bulletin of the Brazilian Mathematical Society, New Series 33(2), 2002, pp. 213-224. A. Kontorovich and S. Miller Benford's Law, values of L-functions and the 3x+1 Problem, Acta Arithmetica 120 (2005), 269-297. A. Kontorovich and J. Lagarias Stochastic Models for the 3x + 1 and 5x + 1 Problems, in "The Ultimate Challenge: The 3x+1 Problem," AMS 2010. Tao, T. (2019). Almost all orbits of the Collatz map attain almost bounded values. arXiv preprint arXiv:1909.03562. — https://ve42.co/Tao2019 Conway, J. H. (1987). Fractran: A simple universal programming language for arithmetic. In Open problems in Communication and Computation (pp. 4-26). Springer, New York, NY. — https://ve42.co/Conway1987 The Manim Community Developers. (2021). Manim – Mathematical Animation Framework (Version v0.13.1) [Computer software]. https://www.manim.community/ ▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀ Special thanks to Patreon supporters: Alvaro Naranjo, Burt Humburg, Blake Byers, Dumky, Mike Tung, Evgeny Skvortsov, Meekay, Ismail Öncü Usta, Paul Peijzel, Crated Comments, Anna, Mac Malkawi, Michael Schneider, Oleksii Leonov, Jim Osmun, Tyson McDowell, Ludovic Robillard, Jim buckmaster, fanime96, Juan Benet, Ruslan Khroma, Robert Blum, Richard Sundvall, Lee Redden, Vincent, Marinus Kuivenhoven, Alfred Wallace, Arjun Chakroborty, Joar Wandborg, Clayton Greenwell, Pindex, Michael Krugman, Cy 'kkm' K'Nelson, Sam Lutfi, Ron Neal ▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀ Written by Derek Muller, Alex Kontorovich and Petr Lebedev Animation by Ivy Tello, Jonny Hyman, Jesús Enrique Rascón and Mike Radjabov Filmed by Derek Muller and Emily Zhang Edited by Derek Muller SFX by Shaun Clifford Additional video supplied by Getty Images Produced by Derek Muller, Petr Lebedev and Emily Zhang 3d Coral by Vasilis Triantafyllou and Niklas Rosenstein — https://ve42.co/3DCoral Coral visualisation by Algoritmarte — https://ve42.co/Coral
I can't believe I came back to this video but I had a weird dream that the reason this happens is actually not an anomaly at all. It's all based on probability. Because of the rules of the game there is a higher probability of ending up with an even number thus allowing you to divide by 2 thus following the 421 loop down. The 421 loop is like the waterfall of the equation. And the 3 multiplier with the +1 is what makes it possible to begin with. Without using the number 3 specifically and adding 1 to make that number more probable of being an even number this would never happen. The reason multiplying by 3 leads you to getting an even number is because it will allow that number to be an odd number again but then you have to +1 which makes it an even number thus allowing the 421 loop to act as gravity through division. Example: If you had the number 5. If you multiplied 5 by 2 instead of 3 you would get 10 which would then become 11. This would repeat over and over and always be an odd number for ever multiplying itself. But because if the 3 the 5 would become 15 which would then become 16 and you would divide to 8 then to 4 then to 2 then to 1. Waterfall affect. All even numbers that end in 0 like 10, 20, 30, when divide by 2 will lead you slowly towards 5 which will lead you to the 6. And the 6 will lead you slowly towards an 8 and the 8 will lead you to the 421. The 3x+1 is like the sails on a ship, the ÷2 is like gravity, the 5 is like the captain of the ship, the 6 is like the the dude on the top of the ship with the binoculares, and the 8 is like the current of water that leads you to the edge of the waterfall and the 421 is the waterfall. Yes I know it's a lot but just try and think about this while doing the math. You will see the patterns. It's just the way the numbers work when multiplying by 3 and then adding 1. THE 421 WATERFALL. Alright I'm done thinking about this forever. I may end up being a lunatic if I think about it anymore and I swear to God if I have another dream about it I'm going to a psychiatrist. Lol
I didn't read your comment because it's too long, but I admire your typing skills....
Let me know if you guys solve this equation, because I think it will lead us on how to make a time machine, perpetual motion, or something 😂
its not that multiplying by 3 and adding 1 is more PROBABLE, it makes it DIRECTLY go to an even number. which halves, and can be either an even number again or an odd number again. but it will always be an even number, which can lead directly to 4>2>1 once it enters 8.
@Dinkhu fr
Why do I have a feeling that 1111 and 4444 have something to do with this? Can you help me? I'm not much of a mathematician.. I'm better at spotting coincidence of images, numbers, and concepts.
Are there any practical or theoretical purposes that solving this problem would achieve? (Not a hostile question, I’m legitimately just curious)
@Brage Sørensen the Reiman Zeta function may have tied to the problem. Who knows?
@Abbadon But this is a made up equation. It has no calculating basis to help any logical problems of any purpose.
If the conjecture is false then there will be super large numbers not obeying the law tested by computers so far. Kind of different world
@Alberto Me It seems that ten is everyone’s go to. What if It never wears a golden crown?
even though it seems irrelevant it wwill probably tie into something else in the future because thats just how vast maths and science are, for example think about the reimann hypothesis, it describes the probablities and distributions of prime numbers on the number line but it also happens to describe the distribution of heat across a body, similarly the graph of the cosh function is identical to the cross ssection of the shape formed when a bubble is made between two hoops
Playing around with this problem makes me strongly believe that all positive integers will reach 1 and that the 4-2-1 loop is the only loop. The 4-2-1 loop might just be a quirk of the problem. I think this problem will be proven eventually.
@Twave no it has not
@Mickey P There are plenty of proofs out there. "Widely" repeatedly verified proofs are zero in number.
@Twaveno??
No one's talking about the fact that there is secretly another loop
THE 16 and 8 one
Basically any natural number other than 16 or 8 always ends in 16 , don't believe me here's one example .
7×3=21+1=22÷2=11×3+1=34÷2=17×3+1=52÷2=26÷2=13×3+1=40 ÷2=20÷2=10÷2=5×3+1 = 16
See any number n and after a p number of steps will hit 16
N after (p) number of steps =16
Okay this is not a loop but still the '16,8' loop you will have to go to 16 and 8 before getting to the the 4 2 1 loop
Ok, I'd like to see how this problem holds up to non integer numbers, or even if it can be modified to include complex numbers.
Well the whole even/odd thing wouldn't work for non-integers
@breadsticck correct
Mad respect to the animators here. That must've been a lot of work.
i wonder if there an effective way to animate like this, or we just do it manually?
@Michał Giedrojć I just wrote a pgm to calculate this easily soo i think they wouldve done the same
Its pretty basic so im pretty sure anyone will be able do it
Thank you for saying animators instead of just “the editor”
Bro you could do the add first and then times by 3 3 ez
It just looks like the division by 2 of even number's has the statistical advantage over 3x + 1.
A division by 2 will always follow after 3x + 1, however that is not the case after a division by 2. A number like 1024 can be divided all the way to 1 without allowing 3x + 1 to come into play
This would be the same case as for 5x + 1, except the end loop would be slightly longer.
1 - 6 - 3 - 16 - 8 - 4 - 2 - 1
7x + 1 end loop would be
1 - 8 - 4 - 2 - 1
This is explained at 8:30
For large numbers, the geometric mean between two odd numbers are 0.75.
That's the crux of the problem, how often does 3x+1 yield an even number that is divisible by 2 or 4 or 16 or 32 or...?
If 3x +1 = 2y { y being half the resultant even number}
3x = 2y -1
This can only be true if x is an odd number.
Likewise, if 3x+1 = (#)y
3x = (#) y -1
This can only be true if x is an odd number.
3x + 1 does always yield an even number, while increasing the preceding number 3 fold+,
followed by a halving.
Apparently, this arithmetic process generates a significantly higher percentage of even numbers
than odd numbers (a "statistical advantage").
Odd number > (3x - 1) -> Even number > (3x-1)/2 [50%?]
Even number > (x/2) -> Even number > (x/2) [25%?]
-> Odd number > (3x-1) [25%?]
Tl:Dr 3x+1 has infinite numbers that x can be, however it loops back into the 4,2,1 cycle, which means it isn't possible to solve. For further explanation, 3(10)+1 = 31 odd, 3(31)+1 =92 even 92÷2 =47 odd 3(47)+1 =142 even 142÷2= 71, this will continue to 1
Basically any number that eventually reaches a power of two lands back on 1, so theoretically any number that doesn't land up on the power of 2 when multiplied by 3 could be the solution. (my way of thinking, no idea if this is correct).
You're correct in identifying that a power of 2 always lands back on 1. However just finding a number that, when multiplied by 3 (and I think you meant to say when multiplied by 3 then added to 1) is not a power of 2 is not sufficient, because that does not guarantee that the rest of the sequence does not eventually land on a power of 2. Take 3 as a counterexample: 3*3 + 1 = 10 which is not a power of 2, however the sequence for 3 goes 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1.
If you prove that any seed number will go to one unless it goes into a loop, have you then partly solved the problem?
Sure, then you've proved it can't increase indefinitely.
Fun fact: We are not mathematicians but we got interested by this.
What we got now is I lost touch with reality this is not a Democrat liberal Party this is a fascist Socialist Party is what this is it is not going to be tolerated. Read Ezekiel chapter 45 and just read the chat read the verses and as you read it notice what chapters 45 what number president was Donald Trump when he was in office 45 look at the sevens in his name and it was his birth date and everything when he was getting out when he was going to be inaugurated all this it says in Ezekiel I look for a man or woman that would stand at the gate and it will build a wall to keep naysayers out that's prophesized Trump walked through a gate the door when trade Center's got hit and get witnessed it and you know he said he seen more people from the Middle East on the in the roads on the sidewalks on their knees and praying and saluting and pleading to Allah because this happened I guess if I'd have seen four or five people hunting people doing that I wouldn't I don't know it it hurt doesn't have to be like that nobody needs to die until the Lord calls them he's never late.
đöń'ț håvĕ åńý šýmpåțhý öŕ ůńđĕŕšțåńđïńg föŕ čhïńå! ï ůšĕđ țö bĕ å čhïńĕšĕ åńđ ï wåńț țö țĕłł föłłöwïńgš țö åłł öf ýöů !:
țhĕ čhïńĕšĕ håvĕ ńöț pŕöđůčĕđ åńý ńåțůŕåł ščïĕńčĕ, šöčïåł ščïĕńčĕ, öŕ hůmåńïšțïč țhöůghț ïń țhĕïŕ 3,000-ýĕåŕ hïšțöŕý.
čhïńĕšĕ čůłțůŕĕ ïš ńåŕčïššïšțïč ńöńšĕńšĕ, fůłł öf åńțï-hůmåńŕïghțš åńđ öțhĕŕ čŕåp.
țhĕ țwö bïg ŕůbbïšh țhåț čhïńå håš čöńțŕïbůțĕđ țö țhĕ wöŕłđ (ńöț łïmïțĕđ țö țhïš): čhïńĕšĕ čůłțůŕĕ (čöńfůčïůš țhöůghț, čhïńĕšĕ mĕđïčïńĕ, ĕțč.), țhĕ čömmůńïšț Påŕțý öf čhïńå (ččP).
țhĕ čhïńĕšĕ pĕöpłĕ håvĕ ńö hůmåń ŕïghțš ïđĕåš ïń țhĕïŕ ńåțůŕĕ. čhïńå håš pŕåčțïčĕđ țĕŕŕöŕïšm đömĕšțïčåłłý föŕ țhöůšåńđš öf ýĕåŕš.
țhïš ńåțïöń håš båŕŕïĕŕš țö țhĕ čögńïțïöń țö țhĕ wöŕłđ.
čhïńĕšĕ čůłțůŕĕ, ïš å bïțčh-łïkĕ čůłțůŕĕ !
ïń țhĕïŕ 3,000-ýĕåŕ hïšțöŕý, țhĕý půț țhĕmšĕłvĕš țö đĕåțh ågåïń åńđ ågåïń, đůĕ țö țhĕ łåčk öf čöŕŕĕčț čögńïțïöńš åńđ mĕțhöđš.
ïń țhĕïŕ hïšțöŕý, țhĕïŕ öńłý šöłůțïöń țö šöłvĕ țhĕïŕ šöčïåł pŕöbłĕmš ïš țö måkĕ ĕvĕŕýțhïńg å mĕšš åńđ đĕšțŕöý, țhĕń šțåŕț övĕŕ wïțh țhĕïŕ čöčkŕöåčh-łïkĕ fĕčůńđïțý.
ĕvĕŕýțhïńg ïń țhïš čöůńțŕý ïš łïĕ, åńđ šhåmĕłĕššńĕšš, țhĕý čömpłĕțĕłý łåčk țhĕ đïšțïńčțïöń bĕțwĕĕń țŕůțh åńđ fåłšĕhööđ.
Whåț pĕöpłĕ ïń țhïš čöůńțŕý đö ïš åłwåýš šqůĕĕžïńg țhĕ ńĕxț gĕńĕŕåțïöń, šqůĕĕžïńg țhĕ böțțöm.
țhĕ čhïńĕšĕ kĕĕp pŕĕțĕńđïńg țhåț țhĕý åŕĕ å ńöŕmåł čöůńțŕý whĕń ïń fåčț țhĕý åŕĕ å šýšțĕmïčåłłý ĕvïł čöůńțŕý.
țhĕ čhïńĕšĕ håvĕ bĕĕń țŕĕåțïńg țhĕ whöłĕ wöŕłđ åš fööłš, kĕĕp šțĕåłïńg țĕčhńöłögý, måŕkĕțš åńđ jöbš wöŕłđ-wïđĕłý.
țhĕ čhïńĕšĕ åŕĕ å håŕđ-wöŕkïńg ńåțïöń, håŕđ-wöŕkïńg țö fłööđ țhĕ wöŕłđ wïțh țhĕïŕ čůłțůŕåł åńđ ïđĕöłögïčåł ŕůbbïšh.
ĕvĕŕý đåý ïń țhïš čöůńțŕý måkĕš mĕ fĕĕł đïšgĕšțĕđ!
Math is hard. Stop teaching it in schools.
@Subrina Campbell this is cuz u look math the most wrong way.. its not about solving.. its about where do nature use it-what can i see trough it .. what does it want to teach me ? ; have fun
same lol
The "problem" is that X assumes consistency. And assumption is the basis of statistics. This goes to show that Paul Erdos is right.....mankind's mathematic concepts are far too simple to explain the nuance that forms our reality. We can only simulate it through the law of large numbers which is close but not accurate.
My gut feeling says that there must be a way to logically rule out different possibilities other than 4, 2, 1 loop
I've noticed something, I've put in extremely big numbers into a 3x+1 grapher and they always seem to go 5, 16, 8, 4, 2, 1, I'm just confused.
I would guess if we got a big enough number it would always be odd when divided and grow a little bit while divided therefore would be infinite
I’m still trying to figure out what math problem we’re trying to solve.
@Hectic Hector coz it is what it is
❤
@Richard Farmbrough I found 2 numbers that don't. It's either -1/3 or 0. -1/3 ends with 0 and 0 divided by 0 is 0. So if you use 0 or -1/3 you are not in the loop again.
đöń'ț håvĕ åńý šýmpåțhý öŕ ůńđĕŕšțåńđïńg föŕ čhïńå! ï ůšĕđ țö bĕ å čhïńĕšĕ åńđ ï wåńț țö țĕłł föłłöwïńgš țö åłł öf ýöů !:
țhĕ čhïńĕšĕ håvĕ ńöț pŕöđůčĕđ åńý ńåțůŕåł ščïĕńčĕ, šöčïåł ščïĕńčĕ, öŕ hůmåńïšțïč țhöůghț ïń țhĕïŕ 3,000-ýĕåŕ hïšțöŕý.
čhïńĕšĕ čůłțůŕĕ ïš ńåŕčïššïšțïč ńöńšĕńšĕ, fůłł öf åńțï-hůmåńŕïghțš åńđ öțhĕŕ čŕåp.
țhĕ țwö bïg ŕůbbïšh țhåț čhïńå håš čöńțŕïbůțĕđ țö țhĕ wöŕłđ (ńöț łïmïțĕđ țö țhïš): čhïńĕšĕ čůłțůŕĕ (čöńfůčïůš țhöůghț, čhïńĕšĕ mĕđïčïńĕ, ĕțč.), țhĕ čömmůńïšț Påŕțý öf čhïńå (ččP).
țhĕ čhïńĕšĕ pĕöpłĕ håvĕ ńö hůmåń ŕïghțš ïđĕåš ïń țhĕïŕ ńåțůŕĕ. čhïńå håš pŕåčțïčĕđ țĕŕŕöŕïšm đömĕšțïčåłłý föŕ țhöůšåńđš öf ýĕåŕš.
țhïš ńåțïöń håš båŕŕïĕŕš țö țhĕ čögńïțïöń țö țhĕ wöŕłđ.
čhïńĕšĕ čůłțůŕĕ, ïš å bïțčh-łïkĕ čůłțůŕĕ !
ïń țhĕïŕ 3,000-ýĕåŕ hïšțöŕý, țhĕý půț țhĕmšĕłvĕš țö đĕåțh ågåïń åńđ ågåïń, đůĕ țö țhĕ łåčk öf čöŕŕĕčț čögńïțïöńš åńđ mĕțhöđš.
ïń țhĕïŕ hïšțöŕý, țhĕïŕ öńłý šöłůțïöń țö šöłvĕ țhĕïŕ šöčïåł pŕöbłĕmš ïš țö måkĕ ĕvĕŕýțhïńg å mĕšš åńđ đĕšțŕöý, țhĕń šțåŕț övĕŕ wïțh țhĕïŕ čöčkŕöåčh-łïkĕ fĕčůńđïțý.
ĕvĕŕýțhïńg ïń țhïš čöůńțŕý ïš łïĕ, åńđ šhåmĕłĕššńĕšš, țhĕý čömpłĕțĕłý łåčk țhĕ đïšțïńčțïöń bĕțwĕĕń țŕůțh åńđ fåłšĕhööđ.
Whåț pĕöpłĕ ïń țhïš čöůńțŕý đö ïš åłwåýš šqůĕĕžïńg țhĕ ńĕxț gĕńĕŕåțïöń, šqůĕĕžïńg țhĕ böțțöm.
țhĕ čhïńĕšĕ kĕĕp pŕĕțĕńđïńg țhåț țhĕý åŕĕ å ńöŕmåł čöůńțŕý whĕń ïń fåčț țhĕý åŕĕ å šýšțĕmïčåłłý ĕvïł čöůńțŕý.
țhĕ čhïńĕšĕ håvĕ bĕĕń țŕĕåțïńg țhĕ whöłĕ wöŕłđ åš fööłš, kĕĕp šțĕåłïńg țĕčhńöłögý, måŕkĕțš åńđ jöbš wöŕłđ-wïđĕłý.
țhĕ čhïńĕšĕ åŕĕ å håŕđ-wöŕkïńg ńåțïöń, håŕđ-wöŕkïńg țö fłööđ țhĕ wöŕłđ wïțh țhĕïŕ čůłțůŕåł åńđ ïđĕöłögïčåł ŕůbbïšh.
ĕvĕŕý đåý ïń țhïš čöůńțŕý måkĕš mĕ fĕĕł đïšgĕšțĕđ!
@Andreas Philippou deep
Its the mathematical formula for the saying 1 step forward and 2 steps back... it will always at one point or another always lead back to 1 without question and the reasoning for this is because of the fact that every single number no matter to what power is always going to be divisible by the number 1... (yes that includes 0 in this equation because of the +1 included in the equation) tbh the answer is relatively simple because of the fact that every number that we are aware of ends in 0-9 and as long as all numbers 0-9 can be proven to go back to 1 at any given point then it is proven that it is a true law of mathematics... I just don't understand the debate... all numbers lead back to 1 and 1 leads to all numbers...
@Tomas because all numbers start and end by the number 1 it is a simple mathematical law... the basis of all mathematics... and because 1 is the only number that when even divided by itself equals 1... even using your number 3... 3÷3=1... 3 divisible by 3 still equals 1... all numbers will eventually equate to the number 1 once fully broken down
@Abbadon But we don't divide numbers by themselves here. We only divide them sometimes by 2. "all numbers start and end by the number 1" what do you mean 'start and end'. A number is number it doesn't start or end anywhere. Just more nonsense
.
@Abbadon I'm not sure what your argument is here. The fact that all numbers are divisible by 1 is not relevant to the function 3x+1. You also point out that 1 is the only number that when divided by itself equals 1, and then proceed to divide 3 by itself to get 1, acting as if you have proven a point when all you have really done is proven yourself wrong. I also don't quite understand what you mean by numbers equating to 1 once broken down, because a "broken-down" number is still that number. For example, if what you mean is a number being divided by itself equating to 1, that number is being subject to an operation. It is being changed. It is not the same number you started with. Taken at face value, what you are saying basically amounts to "any number x = 1", which anybody with the most basic education in math can tell you is false. I'd suggest taking a step back and reevaluating your own arguments so that you can provide a more coherent review of your points.
video: "professional mathematicians have been working on this problem for years and haven't been able to prove it one way or another"
@Abbadon : "no it's really simple and I figured it out all on my own"
lol
My brother is Michel J Burns. I'm sure he knows . But I was thinking about this and it's simular to a 3:15 min trick used in a video game silent hunter III. If you take 1:05 and multiply it by 3 you get a short cut answer. I'd you time it for 2:10 min you have to divide it by 2/3 just off the top of my head. Am I right? It all depends how you convert the numbers, right?
Isn't it easier to work only with odd numbers ?
Odd multiples of 3 have no odd antecedent.
Also, if an odd number N can be written N = 8k+5 (k integer), then it exists a smaller N'=(N-1)/4 which gives you the same next odd number.
presumably their "brute force" strategy takes these and other shortcuts, but the skipped numbers can still be considered "tested"
What about 0 ? I'm curious to know if it has any effect on this conjecture
0 is a trivial solution. This is not very interesting, so the conjecture only considers positive integers.
It too ends up in the loop,
See, 3×0+1= 0+1=1
Which is odd so
3×1+1=4
Divided by 2 we get 2 them again dividing by 2 it's 1.
So zero is too all the same.
@Pratham Pratap Singh 0 is even...
@suhrit kanumuru Actually I thought that you have to apply the 3x+1 rule first, without any regards as to the nature of the number, whether its odd or even.
Sorry my mistake.
But I did some little research and found out that 0 is not a solution since the 0→0 cycle cannot be entered from outside.
i.e., The generalized Collatz conjecture is the assertion that every integer, under iteration by f, eventually falls into one of the four cycles(the 4-2-1 and the 3 cycles we get with the negative integers)or the cycle 0 → 0. (The 0 → 0 cycle is only included for the sake of completeness.)
0 divided by 2 is 0
That's all you need to know
Everyone here: "...but just a maaaaybe I'll be the one to solve it."
maybe
@The Greek God of Wall Street that thought was bugging me the entire video. i know nothing about computers but i wonder why people just cant run a program untill they find a number that doesnt fit in.
@JR Bros faX coz 0 is even, but if integers aren't counted, then its just 0-0-0-0-0-0-0-0-0-0-0-0-0
Wait he said any number so 0 because 0 time’s everything is 0
I've already solve it
Damn watched this video at lunch, now can't concentrate at work anymore... you are costing my employer valuable time!! 😀
It looks so trivial! Can't the problem somehow be simplified? It seems that even numbers are irrelevant, because if I can prove it for f(n) - up to an uneven number n - I have already proved if or f(n+1), because even numbers get reduced to a smaller uneven number which has already been proven.
So somehow it must be possible to focus on the uneven numbers and the function 3x+1, to somehow prove there is always an upper bound exponential of 2, because any 2 to the power of y gets imediatly reduced to the 4-2-1 loop and proves the theorem.
but ofc it's not that easy, otherwise the thousands of people who have tried and are much smarter than me, wold have found it by now ^^
If you don't know what the Collatz Conjecture is or how it works, go to 0:00 and watch until 1:47.
I have a theory about this whole thing, and if you are curious about it, read until the end, you will have to.
If numbers continue infinitely, we will never be able to physically prove every number to support 3x + 1. Thus, if we can't prove infinite numbers to support 3x + 1, we must assume that the Collatz Conjecture is NOT COMPLETELY TRUE. Now, you can't prove the previous statement to be false because you cannot plug in every number in existence (NUMBERS HAVE NO LIMIT) to see if they can support 3x + 1. It's like trying to prove something to be 100% true without 100% proof. In order to prove something to be 100% unequivocally true, you need 100% of whatever type of evidence that is used. (Frankly, you might not even know that you have 100% of the evidence) In this scenario, this data is comprised of numbers from 0 to infinity. In order to prove the Collatz Conjecture to be true, you need to find 100% of all the numbers in existence so you can plug them into the formula, which is impossible to do. That is why I personally don't believe that the Collatz Conjecture is completely true, though it does resemble some form of logistical pattern, so it might be PARTLY true.
(Try to understand it the best you can, I tried my best to explain it) (Keep replies respectful and God bless! :D)
You don't necessarily need to show every number explicitly which is impossible. There might be other ways to proof it. For example, we also know that there are an infinite number of prime numbers, but we haven't tested every number
What a stupid troll comment lol
I accidentally discovered the negative numbers side by experimenting with different rules for the operations, and I tried 3x-1 instead of 3x+1. I tested this thing with a program, and up to 115 million, all numbers ended up in one of the tree loops.
I think its more reasonable to be multiplied by negative three and removing one -3x-1 considering being on the negative side of the integers but that would not be true because it would put us back on the positive integer side and hence stuck in a loop again
The problem with exploring it this way changes how the equation works so it (if I'm not mistaken) doesn't act in the same way when solving it as the positive integer version of the equation
this is interesting and im only part of the way through but my first observation is its not a 4-2-1 loop, because it could be 8-4-2-1 or 16-8-4-2-1 so the common loop is any number 2^n converges to 1. so the problem is finding all x when 3x+1 = 2^n... then the next step is where i get confused lol... but thats how far i got
It's the 4-2-1 loop, because 1 goes to 4
@Tomas ah makes sense, i missed that part. it still means any number = 2^n ends in a 4-2-1 loop.
This math problem is actually like my trading portfolio, I can start with any number but end at $ 1
This is life, no matter how often you do something odd, you end up as one…
god damn 😂
This seems to be the most devastating global pandemic.
😂😂😂
This is like whar casino algorithm looks like
I can't solve it mathematically, but deductively it's actually quite simple. Given an infinite number of iterations, it's impossible not to get into the loop. There is no instance where the (3x+1) or 3(3x+1) + 1 will ever not be even. At this point, you divide by 2, and reach a number that one has seen in a previous iteration of the problem which is known to get into the 4-2-1 loop. Apply this ad infinitum, and there you go.
"At this point, you divide by 2, and reach a number that one has seen in a previous iteration of the problem which is known to get into the 4-2-1 loop."
You don't though. If you're at, say 11, then (3x + 1) / 2 is 17, so you haven't reached a previously known value - you haven't yet examined 12 through 16.
@J Modified your comment is wrong what he means is as long as you get a number which is known to reach 1, then you don't need to go any further,
17 becomes 52
52-26-13-40-20-10-5-16-8-4-2-1 so say you calculate a big number and get 1024, you don't need to continue since we already know it's follows the pattern to 1. at some point with real big numbers, we'll probably be able to divide once and already know that it will reach 4-2-1 loop
@LESGOSHOOPING Read it again. I understood what he said perfectly, and why it is wrong.
anyone else notice the wave like patterns at 15:40 when looking at the squares?
Lol
54,55 and 73 have more interantions than 27.
INTERANTIONS:
27:111.
54 and 55:112.
73:115.
I got send into orbit right when he guessed my number at 00:30
Whoever created all those graph animations is an absolute master in after effects expressions
appreciate it
Wait, if you do this 3x+1 for 1.5, you will always find an odd number, and will eventually lead to infinity
try create an expression for aah
@sweatyMonkey x is a variable. Maybe in 5th grade you used x as a multiplication symbol but it's a bad habit when you start using variables.
@huskai no. Your wrong. His comment was 110% accurate. Because of Jesus he get his editing skills in after effects.
Wouldn’t it all have something to do with the fact that you can’t double a whole number and get an odd number…therefore you wouldn’t be able to keep an infinite set of numbers going…it will all eventually come back to 1 if your rule is to divide by 2 on evens and create even numbers from any odd numbers…because that’s essentially what you’re doing…
The step up is proportionally larger than the step down.
I would just like to say that obviously it doesn’t work the same for negative numbers. It’s not asking the same question, you’d have to mirror the equation making it 3x-1 to achieve the same result with negative numbers.
They're not identical, but there's no known reason that 3x-1 would have multiple loops while 3x+1 would not.
I love simple problems and the way someone can explain their use in real life. Most of Veritassium videos have clever names, so clever that your hand just itches to press play. But I noticed one repeating pattern too. I am not a mathematician. I loved statistics in College and I understand how everything we have would not be possible without countless people solving problems for millennia.
EVERY TIME video begins simple enough and easy to follow. But about 1/3 of it's starting to get so complicated that a regular person like me has no idea how to catch up and understand. Maybe I'm just slow. But I know I'm not that dumb. Does anyone else have the same problem as me? So I actually almost never ever watched a full video without falling asleep or just getting too bogged down by facts, words, numbers, and other fancy stuff
Please, if anyone experienced the same I would love to hear your side of it.
One other thing that bothers me. So many likes, views, and smart comments. Obviously, lots of people understand this stuff. And I'm happy for them. But it makes me feel a little too far behind. Ohhh. Boo Hoo. Poor me.
The equation 3x+1 is known as the "Collatz Problem" or the "Ulam Conjecture" and is famous because no one has been able to prove its result. The problem requires you to take an integer and, if it is even, divide it by 2 and, if it is odd, multiply it by 3 and add 1. You can repeat this process as many times as you want, and the conjecture states that any integer starting with an integer will eventually lead you to the number 1. Although this has been verified for all integers up to a certain limit, it has yet to be proven for all integers. This makes the problem intriguing for mathematicians.
Example with 100:
100 is Even, so divide by 2.
100/2 = 50 (50 = Even, so divide it by 2.)
50/2 = 25 (25 = Odd, so do 3x+1)
3x25+1 = 76 (76 = Even, so divide it by 2.)
76/2 = 38 (38 = Even, so divide it by 2.)
38/2 = 19 (19 = Odd, so do 3x+1)
3x19+1 = 58 (58 = Even, so divide it by 2.)
58/2 = 29 (29 = Odd, so do 3x+1)
3x29+1 = 88 (88 = Even, so divide it by 2.)
88/2 = 44 (44 = Even, so divide it by 2.)
44/2 = 22 (22 = Even, so divide it by 2.)
22/2 = 11 (11 = Odd, so do 3x+1)
3x11+1 = 34 (34 = Even, so divide it by 2.)
34/2 = 17 (17 = Odd, so do 3x+1)
3x17+1 = 52 (52 = Even, so divide it by 2.)
52/2 = 26 (26 = Even, so divide it by 2.)
26/2 = 13 (13 = Odd, so do 3x+1)
3x13+1 = 40 (40 = Even, so divide it by 2.)
40/2 = 20 (20 = Even, so divide it by 2.)
20/2 = 10 (10 = Even, so divide it by 2.)
10/2 = 5 (5 = Odd, so do 3x+1)
3x5+1 = 16 (16 = Even, so divide it by 2.)
16/2 = 8 (8 = Even, so divide it by 2.)
8/2 = 4 (4 = Even, so divide it by 2.)
4/2 = 2 (2 = Even, so divide it by 2.)
2/2 = 1 (When it reaches 1 and still continues, it will be on a Loop of 1 -> 4 -> 2 -> 1)
Tradução para Brasileiros (Translation for Brazilians):
3x+1 é conhecido como o "Problema de Collatz" ou o "Conjectura de Ulam" e é famoso porque ninguém conseguiu provar seu resultado. O problema exige que você tome um número inteiro e, se for par, divida-o por 2 e, se for ímpar, multiplique-o por 3 e some 1. Você pode repetir este processo quantas vezes quiser, e a conjectura afirma que qualquer número inteiro começando com um número inteiro levará você ao número 1. Embora isso tenha sido verificado para todos os números inteiros até um certo limite, ainda não foi provado para todos os números inteiros. Isso torna o problema intrigante para matemáticos.
Exemplo com 100:
100 é Par, então divida por 2.
100/2 = 50 (50 = Par, ou seja, divida-o por 2.)
50/2 = 25 (25 = Impar, ou seja, faça 3x+1)
3x25+1 = 76 (76 = Par, ou seja, divida-o por 2.)
76/2 = 38 (38 = Par, ou seja, divida-o por 2.)
38/2 = 19 (19 = Impar, ou seja, faça 3x+1)
3x19+1 = 58 (58 = Par, ou seja, divida-o por 2.)
58/2 = 29 (29 = Impar, ou seja, faça 3x+1)
3x29+1 = 88 (88 = Par, ou seja, divida-o por 2.)
88/2 = 44 (44 = Par, ou seja, divida-o por 2.)
44/2 = 22 (22 = Par, ou seja, divida-o por 2.)
22/2 = 11 (11 = Impar, ou seja, faça 3x+1)
3x11+1 = 34 (34 = Par, ou seja, divida-o por 2.)
34/2 = 17 (17 = Impar, ou seja, faça 3x+1)
3x17+1 = 52 (52 = Par, ou seja, divida-o por 2.)
52/2 = 26 (26 = Par, ou seja, divida-o por 2.)
26/2 = 13 (13 = Impar, ou seja, faça 3x+1)
3x13+1 = 40 (40 = Par, ou seja, divida-o por 2.)
40/2 = 20 (20 = Par, ou seja, divida-o por 2.)
20/2 = 10 (10 = Par, ou seja, divida-o por 2.)
10/2 = 5 (5 = Impar, ou seja, faça 3x+1)
3x5+1 = 16 (16 = Par, ou seja, divida-o por 2.)
16/2 = 8 (8 = Par, ou seja, divida-o por 2.)
8/2 = 4 (4 = Par, ou seja, divida-o por 2.)
4/2 = 2 (2 = Par, ou seja, divida-o por 2.)
2/2 = 1 (Quando chega em 1 e mesmo assim continuar, ficará em um Loop de 1 -> 4 -> 2 ->1)
A big shoutout ot the graphics department for making this 100% more understandable!
TURN TO THE LORD JESUS CHRIST BEFORE ITS TOO LATE, GIVE YOUR LIFE TO HIM AND START WALKING IN OBEDIENCE, WITHSTANDING FROM ALL SIN AND WICKEDNESS, JESUS SAID THE PATH TO HEAVEN IS HARD AND NARROW, AND FEW FIND IT. MATTHEW 7:13-14, HEBREWS 5:9, JOHN 14:15, MATTHEW 7:21-26, 1ST CORINTHIANS 6:9-10, JOHN 3:16-21, JOHN 10:7-8, MATTHEW 10:26, AND LUKE 13:5. GOD BLESS YOU ALL.
@myUserName thanks
no problem
@icebreaker900 Broad is the path that leads to 1. Narrow the counter examples to this conjecture.
@Anndy Arguedo lol
I was recently watching a video on the Cantor Set, particularly how when trying to figure out a pattern, it was helpful to represent it in base 3. I've also started learning about computer science and how it stores in binary, base 2. I found it interesting how typically (101 = 1x100 plus 1x1) and in binary (101 = 1x2^2 plus 1x2^0). Maybe binary will make it clearer as to why numbers fall back to 1.
I came up with an answer 0 is classified as a number an even number and 0 divided by 2 will always go back to 0🤓
3x+1, divide by 3, x+1/3. Then divide x by 1/3 and you get. x//1/3 + 0 then multiplye by 1/3 and since the other number is 0, the x is left alone. Thats all lol
Did you watch the vid
My answer is correct! 3×+1=4 here's s how, so u multiply the 3 and 1 (3×1) equals 3 and then u add 1 (+1) then the answer is 4!
Mathematicians: Dont waste your time on this problem
20.7 million people: YES
We need to work on practical problems that solve mankind's various problems such as air water and ground pollution, developing cleaner energy, food water and resource supply and more equitable distribution. Having solved these problems then we can move on to developing Warp technology to open up the final frontier.
Search about Abhijeetbyte Collatz Conjecture GitHub
😎🤣🤣👍👍..... Can't share Links on YouTube comments
26million*
@Frank Chary imagine if we do figure out warp technology and they actually call it warp technology lol
@David Medina I just tested few billion numbers using C, but i don't have long enough data type to test more. So i'm probably gonna have to implement longer data type to test more numbers. I'll keep y'all updated :D
Bro knew I chose 7. And he about to blow my mind with this
How I solve 3x+1 is just by using bidmas, brackets, indices, division, multiplication, addition and subtraction. As you can see the multiplication comes before the addition in bidmas, so I multiply the 3 by the 1 resulting in 3.
It’s just 3x+1. Like 8x-5 or 4x+9 or 7x-3. No I didn’t watch the video in case you’re
wondering.
I haven't looked deeply into this problem and although I think it would be interesting to see all the covering aspects...But in my opinion it is fundamental to go back to the concept of mathematics and its representation in reality..Just like this video has protrayed the Collatz Conjecture as a height representation and also the frequency of other sequences such as the Fibonacci and golden ratio in nature, taking into account that reality and nature is made of positive integers I feel that through this conjecture we are not able to find 2 consecutive odd numbers that follow each others (I haven't profoundly performed it on large numbers) and thus we are not able to perform two consecutive 3x+1 consecutively...I do not claim anything but as they spoke in the video trying to find a counter argument will be the solution to this problem and not continue to feed its correctness.
Oh my god, this poor animator. That is a serious amount of dedication. Looks fantastic!
He needs to be paid every single day $100,000 heh
@Llama Man no, you're amazing
@Lucky The Luckless Wolf I know I am
@Timmyrbx how do u know it’s a “he”
@Mehtab Ghumakkad Maybe I'll work with this someday
Python fascinates me everyday
3n + 1 always results in an even number since any odd number x3 is odd, add 1 becomes even.
An even number /2 has a 50% chance of being even. (I think)
It seems to me like the numbers are just bouncing around with a higher probability of being even until you eventually land on a power of 2.
Once you do 4,2,1 is guaranteed.
I don't think this is by any means a rigorous proof but are you not just repeating the process over and over until you get a binary number?
I'm no mathematician though!
what about decimals...? they are real & positive numbers, it just does not fall into the even/uneven category
Integers can only be whole real numbers
@Jason Lian Integers are not mentioned.
@Edward Clark Yes they are, he said a positive integer. Decimals are not integers
Isn’t it like 4x but we don’t know the value of x. I might be wrong😅 but it seems like it 4x was my answer when I saw the thumbnail
Try watching the video, the thumbnail is not the complete problem. In fact the "3x+1" in the thumbnail is not a problem at all as it is not something that can be solved.
The Collatz conjecture is a mathematical problem that involves iteratively applying a certain function to a positive integer, and then repeating this process with the resulting value. The conjecture is that no matter what positive integer you start with, you will always eventually reach the number 1 through this process.
Here's how the process works:
Take a positive integer n.
If n is even, divide it by 2 to get n/2. If n is odd, multiply it by 3 and add 1 to get 3n + 1.
Repeat step 2 with the resulting value of n.
For example, if we start with the number 6, the sequence of values would be 6, 3, 10, 5, 16, 8, 4, 2, 1. If we start with the number 7, the sequence of values would be 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.
The Collatz conjecture has been tested extensively and has not yet been proven or disproven, so it is still considered a conjecture. It is also known as the "3n + 1 conjecture" or the "Ulam conjecture," after the mathematician Stanislaw Ulam, who is credited with introducing it.
Cosmic Nomad - 2022-01-08
I absolutely love how mathematicians always find the most random things to debate over!
neil scott - 2022-12-14
Has anyone applied quatum mechanics to the problem
The Skull Emoji - 2022-12-24
@Christ Loen Maybe, but many mathematicians tell you not to work on it because it's a waste of time
Larry Stevens - 2022-12-24
Ha, I see what you did there.
Guest13 - 2022-12-31
@Orezio Pancrazio do you consider art a waste of time? if yes, what do you consider a use of time?
OTC GAMER - 2022-12-31
They have too much time
So much that they just make some things to destroy future generations