My honest attempt at the Collatz Conjecture | Full movie
Вставка
- Опубліковано 4 чер 2024
- Video too long? Read the paper instead: drive.google.com/file/d/1iydE...
The Collatz Tree can be made with three tiles, and this has some interesting consequences
Help me pay me student debt (please...): / highlyentropicmind
For one time donations:
Buy me a coffee (best option): www.buymeacoffee.com/highlyentrc
Ko-fi: ko-fi.com/highlyentropicmind
Paypal: paypal.me/HighlyEntropicMind?...
----------------------------------------------------------------------------------------------------------------------
GitHub: github.com/Frigorifico9/Const...
Want to download the design of the tiles?: drive.google.com/drive/folder...
----------------------------------------------------------------------------------------------------------------------
Special thanks to all my patreons, but specially to:
OtiumOtiosum, Valerie Hyde, T Highfill, Ryan Roberts, Carlo Fazioli and Cookie_T and Azure Azalia
----------------------------------------------------------------------------------------------------------------------
Credits:
Script, Host, Filming, Directing, Necromancy, Editing, Production: Fernando Franco Félix
Animations and graphics: Oscar Flores and Fernando Franco Félix
The song used thoughout the video is "Song Of The Sirens" by Gaymaranda, and I paid her for the permission to use it: • Song Of The Sirens
Sources:
I purposefully did not use any
----------------------------------------------------------------------------------------------------------------------
Chapters:
00:00 - Intro to the Collatz Conjecture
05:46 - Analayzing the Tree
12:15 - Super Awesome Tiles
15:00 - Analyzing the tiles
17:49 - Infinite Stark nodes
23:46 - Lannister and Targaryen streaks
24:44 - Multi-streak drifting
27:34 - Collatz and chaos
28:26 - First ending of the video
28:56 - Infinite constellations
34:22 - Don't be a crakpot
37:07 - Scott Aaronson's guide to being wrong
40:31 - Cyclical constellations
42:46 - My open problem
43:21 - Negative cycles
45:59 - Conclusion
47:00 - Second ending of the video
47:24 - The Super Awesome Tree
----------------------------------------------------------------------------------------------------------------------
Attributions for graphical resources not stated in the video itself:
For the pciture of Lothar Collatz:
Author: Konrad Jacobs, used under ca-sa 2.0 Germany
"It is very easy to compete against dead people because they can't earn any more points" quote of the year, lmao
How can I copy this comment?
Unfortunately the subtitles say 'deaf people' rather than 'dead people' which has the completely unintended effect of slighting the very people using said subtitles. The original quote is great though.
@@MrDannyDetail oh, the irony!🤣
“Sometimes you have to do side quests to advance the campaign” - I love this!
Several years ago I dreamt I had solved the conjecture.
I have never been so mad about waking up.
Because it is solved. I did it, others did it. (look my other comment i just posted). TL;DR: The only reason you dont know about it, because Khazarian Maffia Says: "this hoax must live to keep idiots buzy. and no, we do not give the promised million." There. Now you know. Free yourself.
I was messing around and ‘proved’ it once. Was so disappointed when I was checking everything and I was off by a +1 somewhere along the way.
@@evanhagen7084 Lol, that must've been mortifying.
@@evanhagen7084 did you (ups, he) make a mistake? Collatz concecture is biggest hoax (that it is not proven)? Satanists are fucking with ppl.
Whether or not your idea directly provides a proof of the Collatz Conjecture.... by God, you achieved your goal. This video was so thoroughly thought provoking, and you did an AMAZING public service by teaching proper skepticism. That alone earns this video a FANTASTIC score in my eyes for that, but the actual progress you made in digging through the Collatz tree was FANTASTIC as well!!
I live for these comments
Interesting things happen when you allow negative numbers. There is exactly one cycle of
Length 1: 0
Length 2: -2 -1
Length 3: 4 2 1
Length* 4: -2 -1 -2 -1
Length 5: -20 -10 -5 -14 -7
There is no reason why this uniqueness should continue, as the number of cases to consider keeps increasing. But are there lengths which have no cycle at all?
*) That's actually a double cover of the Length 2 cycle, there is no other way to make a length 4 cycle. Might be better to say there is no length 4 cycle at all.
Dude, I've been following your channel (and thumbs-up-ing your text posts) for at least a year or two, and rewatching the beginning of this video reminds me why. this is honestly so fucking inspiring, from somebody who is so bored with being alive, you are reminding me why i study math even though i'm not "a success". me encanta a tu "djiutub" video y tambien tu acente me hace sentir comoda jaja, entonces, muchas gracias!!! no puedo esperar hasta tu video QFT!! @@HighlyEntropicMind
@@lexinwonderland5741 Thank you so much, comments like yours motivate me to keep going, similarly to how I seem to be helping you. Keep going, for the joy of learning. Also, just now I'm editing the next episode
Just one question: What do you mean by my "djiutub" video? I'm not sure what you mean
@HighlyEntropicMind LOL sorry i was trying to phonetically write your pronunciation of UA-cam, bc hearing such a familiar accent felt endearing. Anyway im looking forward to your next UA-cam video friend!!! I appreciate the text posts between QFT videos because they let me know you're still around
Regardless of any progress on the Collatz Conjecture, I'm indebted to you for introducing me to the phrase "As though my ass had fallen off."
Saw this comment as he said it, truly a great phrase we all need to start using
i cant believe i read this comment the second that he said that exact phrase lmfao
It's a reference to the book I mention at the start
I loved it because in Portuguese we have the same expression "de cair o cu da bunda" (it is of falling the asshole from the butt).
It means to be surprised, perplexed or even horrified
The story referenced features a boy born with a great golden screw in his belly. His parents don't understand it, so one day he goes out into the world, asking at the big towns, the famous hospitals, the great halls of learning, the many monasteries. Still, not a single person can tell him what it is or why he has it. Finally, he comes before the king, and, prostrating himself before his infinite wisdom he asks: can you explain this golden screw?
The old king is thoughtful, until, with reverence, he reaches into a nearby gilt cabinet, and reveals a great golden screwdriver. He places it against the boy, and it fits the screw perfectly. The boy is overcome by excitement.
The king turns the screw once. Nothing happens. He turns it twice. Still nothing. He turns the screw one last time. And his ass falls off.
The riddle mentioned at the start of the video came when the dad asked what the story meant, and, being nonsense, it didn't, that's the whole bit. Of course, in the novel as a whole this capacity to seriously apply ones mind to things that are impossible to understand is a very important theme
As soon as you didn't assume my intelligence and described what a graph was, that's when I subscribed. There's nothing more frustrating than trying to follow a video that has gone too deep into whatever topic it's discussing without a proper introduction to the material first. This is a fantastically entertaining and easy to follow video, thank you.
@@sirshendu2e01 It’s a growing trend at least! Sure, if you watch a ton of videos in a row it could get a bit annoying (like when I got really into cellular automata and every video had their own introduction to Conway’s Game of Life), but we have the mystical power of just… skipping those sections if we don’t need them.
Also, for the youtuber, it’s good practice and helps them build the skills of explaining things in general.
I think were things went wrong is when going from series to Infinity. Just because an algebraic manipulation works for each step in the series doesn't mean that, that same manipulation holds when talking about the limit. You can only split a series in parts when those parts do converge and as you said Integer series don't converge so you cant just split them an expect that to still converge
I think you're right here.
I want you to know this is one of my favorite math videos I’ve ever seen. You have, more than anyone I’ve seen, encapsulated not just the beauty but the FUN of math. A+ work my dude, I hope to see more! I’d love more of that same style where you go through your own reasoning
* blushes *
I can't say this any better than @MaxxTosh said it, loved the style of this presentation and am inspired to encourage this kind of thinking in my students every day.
I have been working on this problem independently for years (around 3.5 years at this time). I started in my freshman year of college, and now in my senior year I am still making discoveries.
This problem is so taunting, it feels like the problem is cursed by demons to trap mathematicians into chasing after ghosts.
This is your video game dopamine loop. You are clearing out the same cave in Skyrim, because there will always be a harder boss with better loot.
I've been working on it for about 10 times that long. Don't give up.
I understand now that the Kingkiller Chronicles series has not been finished because it is a riddle with no solution! Patrick Rothfuss has been training us how to think by not releasing the third book! Bravo Rothfuss, for this 5d writing strategy.
Eeevil! Evil is everywhere.
God-tier author
Half-life 3 confirmed?
His ass fell off
When I was a sophomore in high school a handful of years ago, I thought I had solved the Collatz conjecture at three in the morning, called all my friends, and woke up the next morning to realize I was completely wrong.
You indeed made a mistake. You will notice it when you try to apply this technique to some infinite sequences. For example, the arithmetic sequence a(n) = 2n+5, or a(n) = 2*(-1)^n+3. You wrongly assume that there is some "last" number a(inf) and then show that it is impossible to construct an analogous Diophantine equation for it. It does not follow from this that this sequence is not infinite.
Interesting, I want to understand your argument better. If the sequence exists, why can't we construct it's equation?
@@HighlyEntropicMind In short, because what you're thinking of isn't a sequence equation.
Your equations determine two variables that mark the beginning and end of the sequence/constellation. When a constellation is finite, the beginning and the end determine the whole of it. That's why you can call such an equation the "constellation equation". But the non-existence of the equation unfortunately does not prove the non-existence of the constellation.
Problems with constructing the "infinite constellation equation" may suggest that there is no such equation (another thing is that you cannot immediately rule out that there exist some other secret technique that allows you to create such an equation). In this case, you are right - there is no such equation, because there is no number that could be called the end of such a constellation.
@@HighlyEntropicMind Suppose that the sequence looked like a_n = a_0 + n. (I know this doesn't work but if I could find you one that worked I'd have proved the conjecture false so bear with me.) In that case, you wouldn't be able to calculate a value for a_inf, because you'd just get infinity. And that makes sense, a_inf is not a real thing, and it's not part of the sequence. Like here's a question: with the infinite sequence I provided, what is a_inf mod 2? Well, the parity of a_n alternates between 0 and 1 so the limit of a_n mod 2 is undefined.
You should also consider that the existence of an expressible equation at all is unlikely to exist if the conjecture is false because of an infinite constellation. If the infinite constellation repeated, then you could convert it into a cyclic constellation, which could result in a cycle of numbers as well. But if the infinite constellation doesn't repeat, then trying to find a closed form for the nth term seems like trying to find a closed form for the nth digit of pi.
@@HighlyEntropicMindHey, looks like your option 2 happened here after all 😁
@@HighlyEntropicMind I think the most succinct way to phrase @Integer0's point might be that a sequence with a countably infinite number of elements cannot have both a first and a last element. If it has both, then the sequence must either be finite in length or have an uncountably infinite number of elements. (I could be wrong, though. I'm a lowly physicist; a tourist in the mathematician's realm.)
The second part of this video shows the value of community in any kind of research: sometimes you need a fresh pair of eyes to find a mistake or inconsistency in your argument. So instead of trying to answer if you are a crackpot or not it would be better to find somebody else and ask them what they think about it. This is also the essence of supervision and peer review. One really can't approach their own writing in the way you read what other people wrote this is why you really need somebody else to go over it. I am finishing up my Japanese literature PhD disertation and I am appaled at the number of non-sequiturs and spelling mistakes I've done in it! (it's like 300 pages after all).
Join a communist party to help expad the community available to work on such problems.
*appalled 🙂
I like the emphasis on having intellectual humility, while still encouraging the process because it's what leads to learning. That's really the only reason I've ever fiddled with the conjecture. Here's what I came up with: If we suppose a strong induction proof might work on the Collatz conjecture, then we only need to look for starting numbers such that their sequence that follows will eventually dip below the starting number. Thus, it doesn't matter how fast we proceed through the sequence, so let's define an accelerated collatz function.
If n is even, then n=(2^k)*m for odd m and some k. If n is odd, then n=(2^k)*m-1 for odd m and some k. Then the accelerated collatz function is given as follows: f(2^k*m)=m and f(2^k*m-1)=3^k*m-1. This function has some nice properties, like how k is the number of iterations of the standard collatz function (which often immediately divides 3n+1 by 2, since it is guaranteed to be even), and how f(even)=odd and f(odd)=even.
This gets the same result that you got (namely, a method of generating our choice of sequence) but I think it's a bit simpler in my opinion.
Addendum: I should mention an additional nice property. The values of f(even) will contain the limit inferior, so if it does dip below the initial seed, then it will appear as f(even) for some even term in the sequence.
Oh another thing: if you view k sort of like a random variable (not really because N is countable, so has no uniform probability), then k would have expected value 2 given the density of even numbers. Thus, on average we can expect even numbers to scale down by a factor of 4, and odd numbers to scale up by a factor of about 9/4. So on two iterations, we can expect a net scaling of 9/16, which will decrease on average.
So it’s pretty clear that the collatz conjecture should be true for almost all n.
I'm only 13 minutes into the video but I already like it heartily beacuse it seizes what most other math UA-cam videos, however comprehensible and well-made, fail to portray: the feeling of *actually* doing math. It's really all about the exploration, testing and trying out various ideas that may or may not eventually work out at all, about giving things silly and/or beautiful names in the process just for aesthetics and simply to have fun, about looking at the examples of certain behaviours first and only then trying to formalize them, as to find out whether the patterns you think are true do indeed hold and if they tell you something about the problem you're working on... or a completely different one. This is what I'm doing math for, this is why I love it. Huge props to you for being able to share this feeling with the general audience instead of trying to make yourself look perfect and your reasoning polished to avoid criticism. As the community of mathematicians, we should generally accept that making mistakes is a natural part of what we do, and we shall not hesitate to share our attempts, to correct and to be corrected.
And you also helped me to look at the great unsolved problems of mathematics form a different angle. I never really considered actually trying to work on them because I'd never put in words this philosophy of "being able to learn from them forever". I had that feeling inside of me, but it was contradicted by the steel logic of "if you most probably won't be able to solve them, why even bother?"
Now I can see clearly that this is not about finding the actual solution, but about what you can learn while looking for one, so now I fell completely free to give it a try. Thank you for that!
Thank you so much for this comment
pretty interesting video. I tried slapping the collatz conjecture with my very flaccid and wobbly mathematical baseball bat 3 months ago and obviously failed at solving it but the insights i developed throughout this process were actually recognizable in your method, just with a different visualisation. Math is neat
Math: Even once!
Wouldn't a flaccid and wobbly baseball bat just be one of those pool noodle things?
You're right that an infinite sum of nonzero integers diverges, but the sum of the differences may converge, and it looked to me like you split a sum of differences into a difference of sums when you moved to the exponent notation.
It doesn't look like, he did it, although, he's multiplying them to natural logarithms, so it might have an effect. Yet, I don't think since both summation and multiplication are associative operations. But things get tricky when working with infinity so, taking care is good.
This is his mistake.
This is an interesting idea, but I'm having trouble seeing how it could be true, because as far as I know this thing diverges no matter what, look: www.wolframalpha.com/input?i=limit+3%5Ex%2F2%5Ex+as+x+approaches+infinity
@@HighlyEntropicMind But this is exactly the same point where I have gotten doubts as well. Remember: No crackpotting. Several mathematically inclined users are telling you, this is difficult: Do not push it aside, dig in. You have to be absolutely certain, these product series with alternating signs converge absolutely (!) before (!) you rearrange them. Also: the argument that series of integers diverge is like a baseball bat to the head, it does not fit the subtleties of the rest of the collartz conjecture. It would be like a lighthouse light. So everyone else who's come before has been blind? This is unlikely.
"That doesn't mean we should let them have all the fun"
Love it!
"But then i had an idea..."
Pretty much where things always go to hell.
The fractional values of B don't always follow the rules. You can see the sequence go 10->16->13->20.5 and such. If the rules were properly being followed you could never go from an integer to a non-integer. I think the algorithm doesn't work perfectly.
I think the whole point of fractional values of b is that they don't need to follow the Collatz Rules, that's precisely the reason they can work with fractional numbers in the first place
@@falnica but if the algo can produce sequences that are partially collatz, but then deviate (like example given above), we have to ask what constraint on b produces valid collatz sequences, not just integers, as we see that anything can happen. (i.e. some rational input b might produce a sequence of integers that are not collatz nodes, but you would not know unless you check.) So this algo may or may not yield all constellations, and may yield infinite collatz-disguised whatever. The behavior hinges on how we might further parameterize b to obtain desired output of algo. (Is that even possible?)
It does seem true from video that integer b yields a valid collatz sequence which is also a constellation of specified sequence. This is probably useful to identify this family of constellations and work from there.
You are an absolute joy to watch!! So glad I found this video, it was super entertaining, super interesting, and I learned a lot too! Will definitely be watching your other videos.
One of the best videos i've seen in my life. The sheer love for maths here is impressive. I wish all the best for you my friend.
Interesting video. Fer's conjecture and the super awesome tiles seem like great ideas! Fer's conjecture could be even stronger: if the distance between nodes is 2, it's only when the last number is odd (the Lannister tile), although then it's a bit wordy haha.
About infinite constellations: on 30:38 you assume that the sequences alpha(m) and beta(m) have to converge to some alpha and beta. I fail to see why that has to be the case.. Also, on 32:02, you say the potential a_inf is "clearly" not a part of the Collatz tree, but why not? Could it not connect to 1 through a different path? Finally, at 32:50, there is a problem with the exponent trick, because it essentially says that an infinite product of (3/2)s, (3/4)s and (1/4)s cannot converge. This is clearly false, as a product of just (1/4)s and (3/4)s tends to zero. (This obviously doesn't produce a valid infinite constellation, this is just a counterpoint to the exponent trick). I think the problem here is as follows: once you rewrite it as an exponent, the product becomes a sum, and then you rearrange the terms in the sum, grouping all the 3s (ln3s) together and all the 1/2s (-ln2s) together. Rearranging an infinite sum only produces the same result if it converges absolutely, which this one clearly doesn't.
One more thing: on 42:39, isn't GCD(2^w, -3^r) = 1?
Oh, and on the problem you've encountered in the end: it seems like your algorithm doesn't find all solutions to the diophantine equation, just one family of it. Otherwise you would get all the integer solutions by substituting integer values for b. I am not well-versed in diophantine equations myself, but that seems to be the problem?
Thanks for the video!
Fricking loved the video. Your raw passion of this is just so entertaining to listen to. Your discoveries and ideas are incredible to hear. I hope you enjoyed making it and i really hope you make more. You are great man. Contiune the side quests!
I love your videos ❤
Thanks for making these amazing masterpieces
36:21 "Looks like a clear cut case of murder, stabbed in the back with a protractor while writing a 13 page blog post defending his theory."
How tantalizing would it be if the one exception to the Collatz conjecture was somewhere in the Septillions and it was proven that it was the only other set to loop around itself. Insane to think that we can’t say for sure it isn’t there
I just recently learned that if you use 5n+1 instead of 3n+1 you very quickly get to septillions
@@HighlyEntropicMind Interesting I actually tried to get somewhere with this and just got as far as figuring out that an equation for any starting point n to the next point f(n) would be "f(n)=(5n/2+1)(n mod 2) + n/2" which I found by adding together the equations "(3n+1)(n mod 2)" and "(n/2)(n mod 2 + 1) - n(n mod 2)". The reason I bring this up at all is that the 5n+1 looks very similar to my 5n/2+1. Example to show it works: f(1) = (5/2 + 1)(1 mod 2) + 1/2 = 8/2 = 4, f(4) = (4/2 + 1)(4 mod 2) + 4/2 = 4/2 = 2 etc. The reason it works is just because I used the modulus function to get the value of 3n+1 to be 0 when the number is even and used it to make the n/2 term 0 when n is odd.
Idts. It has reached to over powers of 60 where the conjecture is still proven true.
I am in awe! What an amazing video about an amazing topic made by an amazing guy! This really is awesome, thank you for sharing this!
Diophantus: *in an ancient letter* “please release this letter in future proving the collatz conjecture, and disproving Fer… Diophantus: game point, Diophantus out.”
So glad I found your channel, first video of many for me to dive into, thanks for the hard work!
Hands down one of the best amateur maths videos I've ever seen. Great job!
I absolutely love and agree with your approach. It's not the destination, it's the journey. Even if we don't solve the riddles, we still learn something, and further the combined knowledge of humanity.
The geometric approach is also very intriguing
Really excellent video! I just discovered your channel and am loving your energy and ideas. Your introduction motivated me to pursue solutions and/or non math pursuits which may, at first, appear impossible
I love the way these ideas are presented
This video is amazing! You really showed the joy and exicetment math brings to the ones who study it.
Also, you inspired me to give Number theory a shot (i'm writing this as i try to read further into Vinogradov's book) so thank you for that too!
All the numbers that appear in your graphs are a node, since to be in the graph the number has to be one that appears in some Collatz sequence which passes through that number. IMO it's fine to call the nodes that have multiple connections anything you choose, but if you want to differentiate then you could refer to them as something such as Converging Node, as they are the integers where sequences converge.
Or “branch nodes”.
@@cmyk8964 branch*ing* nodes
when I tackled it I ended up going a different direction: I saw that nodes are either 4 mod 18 , 10 mod 18 , or 16 mod 18 , and they go in a cycle in the /2 path: 10->16->4->10 , 10 nodes are the solutions to 3x brahces, aka "dead" branches, 4 nodes are the solutions to when a node needs to be divided by 4 before needing to be multiplied by 3, and 16 nodes are the solutions to when a node needs to be divided by just 2 before needing to be multiplied by 3...
but I had no idea what I was doing and ended up just going in circles as a layman
Well you have proven that some specific sequences of tiles must start further and further away from 0 the longer they get, so those for sure cannot be present in a cycle because a0 cannot stay bounded, but the rest doesn't seem to ask the actual question (like sure you can find a number that's limit of an infinite constellation, but that's not necessarily the only possible new cycle)
Amazing work. I still have yet to finish the video, but your presentation is captivating!
Man I have been working on and off with an unsolved problem for years. Took a break but this has really inspired me. Love the king killer reference
Really enjoyed video. Incredibly brave to toss your ideas out for scrutiny. Value for non-mathematians (like me) are your thought processes and strategies.
...These types of problems can be addictive even when every idea hits a dead-end. :).
I remember my half attempt at this problem. I printed out the series in binary using black and white pixels and looked for patterns. It formed some very nice triangles.
so did I, and I've tried to flip to base 3 system.
Instead of the collatz conjecture, let's look at f(n) = n + 1 instead. This obviously has infinite sequences.
I'll try to follow along and apply your reasoning to this. If the conclusion is reached that no infinite sequences exist then that's a problem.
12:06 If n = a, we'll call it a Unit. (So just every number)
12:55 The Unit Tile is pretty straightforward. It can just connect to another Unit Tile
13:37 We can place Unit Tiles together, for example UUU or UUUU
15:08 For the Unit it would be n + 1 = s, which has solutions n = a and s = a + 1. So that's how you could find any Unit.
20:43 For the Unit it would be a_m = a_0 + m, which has solutions a_0 = b and a_m = b + m. So that's how you could find any constellation.
24:29 We can skip this, because we only have one type of tile. We can already find any constellation. No need to do threading.
31:07 You're saying that these constants should both converge to something, in order for an infinite sequence to exists. But in our case, by looking at the equation a_m = a_0 + m we can see that Alpha does converge to 1, which is good, but Beta does not converge because it is equal to m and so will diverge as m increases. That seems like a contradiction.
But maybe I'm misunderstanding. I'll admit I'm not THAT good at mathematics.
TLDR:
I think the statement at 31:05 is incorrect and that this is why the proof is wrong.
The claim is made that Alpha and Beta should converge, in order for an infinite sequence to exist.
A counter-example to this is the function f(n) = n + 1. Here the formula for any constellation would be a_m = a_0 + m.
So here Beta (equal to m) does not converge as m increases, and yet infinite sequences do exist, which contradicts the original claim.
yeah, I like this example
@@Mmmm1ch43l thank you
34:01 Is it possible p-adics could be your solution to having integers converge to a finite value?
And this is the reason for Aaronson's test number 7. In fact, this approach has been tried before. Many times.
well shit, now I gotta learn about p-adics I guess
11:50 "it is very easy to compete with dead people because they can't earn any more points"
Maybe I don't understand the ending, but I think that It makes sense that you can extend to the rational numbers, because for every rational number q, the numbers q/2 and 3q+1 are also rational, which means that you can kind of make what I imagine as a 'web' of rational numbers, where every number q has arrows that points to q/2 and 3q+1, and has arrows that point to it from 2q and (q-1)/3. The tree that you describe is sparser because it restricts it to only certain kinds of constellations, but I think the idea still applies. Checking the parity of rational numbers isn't really possible though.
I didn't think I could, watch a 1 hr video after not sleeping the night, at 5 in the morning.
But I am invested. I surely didn't understand some stuff, but I got hooked on a lot of stuff you showcased. The thing at the end is also so great!!!
you make go on
What freaks me out is that I made a disturbingly similar mistake at solving the same problem about a few days before this video aired. My "proof" never pretended to prove the full conjecture, but only that there wouldn't be any cycle other than those already known... and of course it was wrong.
I had a result quite similar to what you get at 42:38 and for some reason believed it meant solutions would correspond to 2^a-3^b = 1 or -1.
Still, I was able to prove a similar result than your "constellation", which has the consequence of proving that there can't be a periodic constellation that doesn't cycle. (so constellation that doesn't cycle must be aperiodic if they exist). Then I found out some guy already had proven this for their thesis (but I am still proud my proof can fit on two pages, is relatively easy to understand, and works for non-collatz sequences as long as they share some of its properties)
I am currently trying to write a paper about it (I probably will leave it as a pre-print, but I can send you the link to it once it starts looking like something and I have an English translation)
You won me in the first few seconds. Love the reference to A Wise Man's Fear.
The math is way over my head but I give the video an A+ for the methodology. Mistake or not, it's a great teaching tool on working through problems.
this video Is probably the best video i've seen this month (possibly this year)
Obviously, there are fractions that if you multiple them by 3 and add 1, enough times, you'll get an integer, but the Collatz Conjecture requires the number to be even or odd. A fraction is neither. I guess if you just checked to see if a number was even or not even, fractions would work.
A new perspective is always welcome.
Don't doubt yourself my man. This is really interesting and the jokes well timed lol
Can you check if all integer solutions that come when you plug in follow the collatz rule, that is to say if you try the sequence that starts with 82, but change b so that it starts with 164, does it follow the collatz rule?, and also can you have different sequences starting from the same value by putting in fractional values?
love your energy -- great video
I love this, really enjoyed going on the journey with you
This was magnificent! Thoroughly entertaining.....it's been a while since I've contributed to my own channel, and I would love to learn Manim and give my own pet problem an "honest attempt" youtube video for the next SoME...this comment is to put that out into the universe so I actually have to do it! Thanks for the knowledge and entertainment
Perfect intro and theme, great video!
This is how research should be done. Every math undergraduate should watch this.
Just a tiny comment, but huge respect for citing the AI art and prompts in the videos. It's in keeping with my university's guidelines on using AI works, and I think it's a good precedent.
There were no images of Diophantus I could use, and I always cite everything
I gave it a go and tried to approach it by finding a loop. Because even numbers just half down to some odd number, I decided to try to find some way to get to each odd and ignore the even numbers which might make such a process very lengthy as the numbers grow. To do this I invented a new operation which I called factor manipulation. When you manipulate the factor of some number (represented by fm_n) it returns the list of prime factors in descending order, with each factor labeled as f_a with a being the order that the factor is in. You can refer to these factors using the operation, and that means you can manipulate each individual factor in an equation however you want. Basically, fm_n(f_a+3) would return the a’th factor of n and add three to it. You can mention more than one factor, multiply factors, and more. The goal was to try to define what happens to the prime factors in a number when the number goes through an addition (the +1 in the conjecture) but unfortunately I don’t have any coding skills, so I’m not able to create a calculator to perform math with my operation. I had fun though! It’s really interesting to see how different people go about solving the same problem.
is the collatz conjecture only in the real number system or can we take it to the p-adic infinity number, because there are many more loops in the p-adic system, especially when considering 1 as a large number, although I think the p-adics can fit in the real number system possibly proving the collatz conjecture wrong for infinite terms?
I dont have enough knowledge to explain why but what you notice with the super awesome tree really strongly makes me think about 10-adic numbers, if I have the energy I should at least look for the 10-adic numbers corresponding to your rational values for b and look for patterns
54:01 "one over eight - integers again" . . . the final number in the sequence is 7979.5, not an integer
Great video! I love your enthusiasm, and your attitude toward learning and being wrong!
Still, it's very interesting
your videos are so good and I love ur character, you seem like a really nice guy
Top 10 things I wish someone said to me in real life
Very thought provoking video, and thank you for uploading a very nice writeup. Regarding the notion that this generalizes the Collatz in a similar way to how the gamma function generalizes the factorial - I think they are quite different, and that the differences are illuminating.
This generalization is not one function, but instead a family of SuperAwesome functions indexed by a streak and a position therein, e.g. These can be viewed as rational (and indeed linear) functions, e.g. S0(a) = 2a+1, S^m0(a)=2^ma+2^m-1, while S1(a) = 3a+2 and S^m[m](a) = 3^ma + 3^m - 1.
As all the tiles correspond ti rational linear functions, any (finite**) iteration of then will be linear with rational coefficients and so can be written SuperAwesome0(b) = (u/v) + (x/y) b. Integer are whenever b=(y/x)(k - v/u). Similarly there is SuperAwesome[n](b) = (t/w)+(c/d)b integer when b=(d/c)(k - t/w).
So a streak that starts and ends at an integer for a specific streak are given by the intersection of the integer spectra of the first and last SuperAwesome functions for that streak. A full Collatz solution would require the intersection if the spectra of all m+1 (assuming m tiles) SuperAwesome functions.
If the rational coefficients can be related to the the streak sequence, then it should be possible to bound what fraction of integral solutions are actually lost when adding one additional step to the streak. Establishing that this proportion is bounded above 0 (or even log(n)/n) could [convincingly?] show that Collatz solutions must vanish.
A similar approach could be taken for cycles as well.
Randomly, at 11:25 your graph with k = 3 and (2k-1)/3 has a typo where it says 5/5 instead of 5/3 :)
Problems probably start happening around 3^34 because that's around 2^53. There are 53 bits of precision in a 64-bit float
Great video. By the way, Diophantus actually used lots of rational numbers, too, but for some reason whoever called the equations Diophantus equations required the solutions to be integers.
Since you mentioned Scott Aaronson, would you be willing give give a "for amateurs" explanation of his Collatz conjecture work?
It sounds super interesting. Somehow it reformulates it in terms of rewriting rules on a string of characters (like your sequence letters?), and culminates in the claim that if matrices with certain properties exist, it proves the conjecture?
I currently don't understand it, and its a shame there isn't a good overview of this more current research on youtube.
There are MANY other videos explaining the Collatz Conjecture in a way that everyone can understand, I mentioned the one from Veritasium in the video, but there's also this one from Numberphile: ua-cam.com/video/5mFpVDpKX70/v-deo.html&ab_channel=Numberphile
Is it possible that there are only a finite number of non-zero integers in the infinite sum? These correspond to sequences of length 0 that don't actually do anything (i.e. map the previous number to a different number) in the sequence. Like when you get to 1 and the only operation possible maps back to 1, that operation's sequence length is 0. I probably misunderstood a lot because I know almost nothing about Diaphantine euqations but that's my quarter of a cent
It seems like when you compose diophantine equations you are only considering solutions that make both original equations integers. You also need the solutions where the first outputs a fraction but the second gets back to being an integer.
This raises the question
10 gives the sequence 10->5->16->8->4->2->1 in the Collatz Tree
since the Collatz Tree is a subset of the Super Awesome tree, 10 should also give the same sequence in the Super Awesome Tree.
However, 10 gives another sequence, namely 10->16->13->20.5->31.75->48.625->...
That means in the Super Awesome Tree, a number can split into two paths, which makes the Super Awesome Tree not a tree anymore.
When you were playing with b as a fraction, I noticed you only got integers when the denominator is 2^m 3^n -- only has prime factors 2 and 3.
Since powers of 2,3 already came up, this may be a hint.
Very good observation
Even that super large one you tested for the long sequence for 82 has a denominator exactly equal to 2^57
But I managed to catch that the shorter sequence didn't actually give integers when b = 1/8
The nodes are integers, but are the intermediate numbers in the branches integers?
This is the first true extension of the Collatz conjecture I've seen (47:24)
when looking on Wikipedia or other Collatz videos (including yours) they usually show fractals that are made through the continuation of the Collatz conjecture. but they have always been "artificially" extended, I don't have enough characters to explain, but the extensions aren't derived from the consequences of the Collatz conjecture. I feel like someone needs to re-do those fractals with these new values you have found... assuming it can be proven that it truly is an extension, by (probably) showing that you can't just get any number through this method.
of course I'm just an amateur, so I don't know that much. I'm really excited the new are that hopefully will be created by this new discovery.
I know what you mean. This extension of the Collatz Conjecture seems natural in the sense that it includes all the sequences we already had, but also many other sequences "in between", like when the Gamma Function generalized the factorials
exactly!
Im curious how you define the relationship between what you call nodes, and a given fraction. Because in the Collatz, this definition is predetermined by the integer itself. 5 will always divide 4 times to 1. 71 will always divide only once to give 107. And then you know what kind of key it is in your system. But with a fraction, there is nothing stopping you from just dividing forever, so how do you define what key belongs to what fraction? Stepping from rationals to irrationals just makes it even worse. You should do a second video just going into more detail of your algorithm and how it is handling fractional inputs. Clearly I missed something. ;)
@@ianallen738 It's really simple
Let's say you give me a constellation like STTLS, then I can calculate the equation of that constellation and I can find all the integer values of a0 that will give me valid Collazt Sequences
But the same formulas could also result in non integer values of a0. Then I could use the formula for a Sartk node with that value of a0, take that result and use it in the formul for two Targaryen nodes, and keep going until I finish the constellation
What I end up with are a bunch of non integer values that nonetheless satisfy the equation for the constellation and they satisfy the equation of each kind of node
In summary, we took the Collatz Rules, we found the behavior they create, and we realized that behavior is not limited to integer numbers. We found the formulas using integer numbers, but they work with any number
at 54:00 when you tried 1/8 as the input for your code, you said "integers", even though the last number was a fraction. I don't know if it changes anything, but I'm just letting you know.
Oh, nice. I stumbled upon similar "number" system when certain polynoms represented in `up` and `down` notation to introduce 3n+1 and /2 operations applied to some random starting number. With random coefficitiens I tried to figure out which of the coefficients are valid solutions so i guess they are kinda isomorphic to GoT notation.
54:18 do you always get integers when b = 1/(2(2n*3m)) (where n&m are integers) ?
That seems to be the case, but I'm not sure how to prove it
Great food for thought
So I guess my next question would be is there a nice formula for all b that result in integer sequences? I don't think the super awesome tree is quite right unless you generalise the decision rule from divisible by 2 to something else, but without some thought I can't think of a generalisation that guarantees that exactly one of the rules applies for any given rational number.
If we extend the conjecture to just all rationals rather than something like all reals, we could differentiate the numbers by the properties of the numerator or denominator in the number's reduced form. Eg. apply the rule based on if the reduced-form numerator of the given number is even or odd.
In the beginning of the conclusion part i imagine some enlightened monk sitting on a stone under some old tree, infront of his class of diciples, lecturing and getting sidetracked to some weird stuff and ending with "...i could be right, and i could be wrong"
a solution for a loop of super awesome tiles would be a solution for a infinite string of super awesome tiles but first we must define some thing that call a cfl of the loop of super awesome tiles it would be the loop but cut off when it repeat; the infinite string of super awesome tiles would be the clf repeat infinitely.
nice
34:15 i think u have it right there why it could converge, infinity minus infinity being an indeterminate form could maybe converge for certain values of mk and mh, basically write it as a ratio of 3^( Σmk) /2^(Σmh), taking logarithm we get a summation of what could possibly be fractions for certain sets of mh and mk
The distance between nodes:
Let 2n + 1 be any odd number. It must be followed by the node 6n + 4 (you can also get there from 12n + 8). As the node is even the next number is 3n + 2, which can't be a node, but it can be even or odd.
Case A) If it is odd, it must be followed by another node (9n + 7).
So assume instead 3n + 2 is even, and thus n is even, say n = 2m. Then 3n + 2 is followed by 3m + 1.
Case B) If m is odd, you could've arrived there from m as well, so 3m + 1 again is a node.
Case C) The last case is when m is even. Then 3m + 1 is odd and thus must be followed by the node 9m + 4 (which can also be reached from 18m + 8).
So in any sequence, between two nodes at most 2 other numbers can occur (and at least one other number must occur).
An example with all cases: 17 52* 26 13 40*(C) 20 10*(B) 5 16*(A) 8 4*(B) 2 1. The nodes are marked with an asterisk. Note that 4 is a node because it follows 8 but also follows 1.
Great video. Congratulations!
There was a rather famous problem of finding an Einstein tile, a single tile that tiles the plane without repetition. Such a tile was found by David Smith in November 2022. Smith was passionate about tilings, but by no means a working mathematician. The paper _An aperiodic monotile_ was published in March 2023.
The number -2 is a Stark node that's followed by infinitely many Stark nodes, all equal to -2.
I think it's fun simply making the tree of the sequences myself and finding patterns that makes growing it slightly easier
Such as the pretty trivial fact that any power of 2 will rush down to 1 as quickly as possible
My attempt at solving it:
In the case of 1, x->3x+1_>3/2 x+ 1/2_>3/4 x + 1/4 x=x Assume m is the multiplier of x, x a solution over 4, and b the addend. The equation would be mx+b=x m would have to be a fraction because a cannot be a negative number. b would have to be (1-m)x.
The only way for a to increase is by using the +1 in 3x+1. The only loop in the Collatz conjecture for 10^20 is 1,4, and 2. Therefore, for b to increase to be (1-m)x (rough estimate, assuming m is 2/3) it would need 2.5*10^19 steps of 3x+1 to reach its goal.
We want a number that repeats 3x+1, x/2 again and again to make b equal to 2.5x10^19. The first number that does a loop of that is 3, doing it one time (3 10 5) Then 7, doing it 2 times (7 22 11 34 17) Then 15, 31, and 63. This follows the formula 2^x+3. So, for a number to have 2.5*10^19 steps of 3x+1 in a row, it would need to be at least higher than 2.5^10^19+3, which is beyond the realm of any computer. This is an overexaggeration, extra 3x+1 steps would come after the start, but even assuming that the start only was a trillionth of the amount of 3x+1 steps, it would still require a number more than the amount of atoms. This is why I think the Collatz conjecture only works for 1, 4, and 2.
Hot diggity dog, dude that ending!! I didn't understand but suddenly it made sense. I never thought that anything but integers could be in the Collatz conjecture, but of course we should have instances where fractions would appear as we are talking about all real numbers up to infinity.
The definition of the collatz conjecture depends on the input number being even or odd. How can you tell if a fraction is even or odd? What Fer seems to have found is some other collatz-like relationship between numbers which is chaotic like collatz (it literally is using the patterns of multiplication and addition taken straight out of collatz, because those are the GOT House nodes), but it's using them on the wrong numbers.
One string of a collatz sequence may say "multiply divide multiply divide divide" because the numbers are odd even odd even even, and Fer's program reads that as SL, but at the end, it seems to be using the SL pattern on a starting number that doesn't follow that pattern in the collatz tree. That's why it goes off the rails into non-integers.
I think just about everyone gets a bit lost and rambley from trying to think about the collatz conjecture though. I'm probably wrong even though I think I'm right talking to you. At least its exercise for the brain.
Well done nerd-sniping me back into thinking about the Collatz Conjecture.
I'm curious what would happen is we focus on special set of hypothetical numbers; and try to prove it from there:
My thought-process:
*IF* a sequence always grows and doesn't return to 1 *OR* if a particular sequence ends in a cycle other than 1; it *HAS* to have a *lowest number* that isn't 1. (right?)
What properties would this lowest Collatz number have? If we can prove this number doesn't exist or can't exist, we will prove the conjecture.
I've done some (high school level) math and came to the conclusion this lowest number *has* to be in the form 12*n + 7. Other numbers have numbers following it that are smaller, or
predecessors that are smaller.
But what else can we find about this hypothetical number? Are there more limitations? This seems like a novel way to prove Collatz.
This could be plugged into your equations to determine any patterns that could lead to this lowest number, and which structure is above it?
This seems to be one of the general directions that research is going, in terms of how powers of 2 and powers of 3 relate to each other. The hard part is that given a complex enough sequence of ups and downs, it can be difficult to tell whether or not your final answer is smaller than what you started with because the +1 part of the 3n+1 step can snowball after enough steps. You can still bound it by an exponential, but theres not much beyond that without knowing exactly how close powers of 2 and powers of 3 fit together
There are techniques which enable you to extract the beahviour of expansion of generating functions. Perhaphs, by reformulating the Collatz Conjecture into combinatorial problem (your structures) could give us better insights and perhaps, by approximating its generating function at infinity could give you some hints whether there is or is not infinite tiling of yours.
I did some very brief playing around with this after being inspired by your video, one thing I found is that for each odd number that increases (4n-1), the next odd number they arrive at increases by one each time, (for example, 3 -> 10 -> 5, which is 1 odd number away, 7 -> 22 -> 11, which is 2 odd numbers and so on). Since those numbers are 2, 4, 6, 8... apart, only half of those numbers will go to another odd number that increases again (only multiples of 4 apart). Multiples of 8 will hit 2 increasing odd numbers, and so on.
Fascinating, you may be onto something here. This could be useful to try to prove that we can reach ever number starting from 1
4n-1 numbers increase in a way where they only hit an even number once before returning to an odd, rather than running through multiple even numbers before coming back to an odd.
I tested some options for n up to 20, and found that the more powers of 2 in n's factors, the more times it would result in another 4n-1 number when run through the collatz machine.
My conclusion is that the number (2^n)-1 where n is very large, would be a great candidate for a 4n-1 increased odd number
The math works like this:
(2^n) - 1 -> 3(2^n) - 2 -> 3(2^n-1) - 1
We've basically traded one of our powers of 2 for a power of 3, and our resultant number is still of the form 4n-1, so it will undergo the same process, turning powers of 2 into powers of 3. With many powers of 2, you can get a very high number of the form 3^n - 1.
Interested to see if anything comes out of this.
This is absolutely an amazing video and really inspiring. I hope the first 5 minuted of your speech about why we should try doing thr impossible on our own goed viral. Think more mathematicisns and physicist needs to think like that.
Finally, a near-miss that won't be named after Parker.
25:00 - Interesting insights so far, and maybe it does point the way towards a formal solution, but maybe we need to stretch the definition of “solution.” The Conjecture basically says that you’ll always run into a power of two and that there are no closed loops. Both have to be proved to satisfy the conjecture. Assuming that loops could be disproved, the existence of a starting number that never resolved to a power of two would imply an infinite number of starting points that never resolve to a power of two, and therefore an infinite number of independent galaxies or regimes of integers that never coincide. You’d perhaps discover a new class of number. Approaching Collatz as “purporting to seed an infinity of disparate and non-intersecting infinities” might “prove” that doubting the conjecture is absurd. On the other hand, if “any sequence” of connections is possible, one which never intersects another might be out there. It sounds like your technique to find any pattern would be unable to pick such a number, therefore that number lies outside of the entire space of the integers.
Considering your whole expansion idea I came up with this
The whole nodes thing has to be revised because rationals make things a little messy
Two options:
A) We remove all restrictions and just consider our tree as a special kind of set with operations (I have no idea what they're called) instead
Our operations must be N := {2N, N/2, 3N + 1, (N - 1)/3}
And we know 1 is an element
Pretty easy to see the only numbers that could possibly be in the set are rationals with denominators of the form 2^M * 3^L
As an example
8 could branch into 16 or 7/3 above
And it could branch into 4 or 25 below
Question A: are all rationals of that form in the set?
(I just realized this could be a group of operations acting on the number 1. So if I call 2N D, 3N + 1 T, and their inverses D' and T', then our group is the set of all possible strings of D, D', T, and T'. Possible that absolute cycles don't exist that aren't just a concatenation of [string][string']. DDT' for example takes 1 to 1 but it doesn't take 2 to 2)
B) Impose a restriction that applies to rationals as well
I first thought of the odd-even rule on the numerator, and that restricts us to the set of all numbers with denominator 3^M.
Example: 10/27 => 5/27 => 14/9 => 7/9 => 10/3 => 5/3 => 6 => ...
Our stark, lannister and tangaryen nodes still apply here!
10/27 would have a sequence of SSSSTT
28/27 would have a sequence of LTLLS
Unfortunately this does mean that any number can be a node since by previous example one of the nodes we ran into was 2
Question B: Do all rational numbers of that form reach 1?
I like your ideas, but I don't see why nodes are a problem. You can just calculate the "a" for each node in the sequence and from there just calculate the value of each node. My code does that, it's the function "getEachNode"
I belive the "Super Awsome Tree" is not an extension of the Collatz tree. You just generate squences with specific tiles applied, regardless if the numbers are odd or even.