The Collatz Conjecture... but in Binary

Let me explain the “3x+1” mystery (the Collatz conjecture) and show some cool shortcuts that happen if we interpret it in binary, as well as some extensions like including negative numbers!
Make sure you're also tuned in to my ‪@Domotro‬ channel for my livestreams and other bonus content!
This episode was filmed by Carlo Trappenberg (plus a few scenes filmed by Jonathan Ruchlis). It was directed and edited by me (Domotro) and I also made the music in the soundtracks. And here's a link to the public domain fractal image I included near the end:
en.wikipedia.o...
A small correction: near the end, when showing the negative number patterns, I had 1-2-4 written above those as an example of the positive loop, but 1-2-4 is in backwards order with the way I drew the arrows (arrows+error=errows?). That was meant to refer to the normal 4-2-1 loop (which could also be written as 1-4-2 or as 2-1-4) that was in the earlier portions of the episode.... There's also a few little typos: I misspelled "result" as "resut" one one whiteboard, and I accidentally typed "and my" twice on one of the credits whiteboards. Whoops.
Special thanks to Evan Clark and to all of my Patreon supporters:
Max, George Carozzi, Peter Offut, Tybie Fitzhugh, Henry Spencer, Mitch Harding, YbabFlow, Joseph Rissler, Plenty W, Quinn Moyer, Julius 420, Philip Rogers, Ilmori Fajt, Brandon, August Taub, Ira Sanborn, Matthew Chudleigh, Cornelis Van Der Bent, Craig Butz, Mark S, Thorbjorn M H, Mathias Ermatinger, Edward Clarke, and Christopher Masto, Joshua S, Joost Doesberg, Adam, Chris Reisenbichler, Stan Seibert, Izeck, Beugul, OmegaRogue, Florian, William Hawkes, Michael Friemann, Claudio Fanelli, The Green Way, Julian Zassenhaus, Bailey Douglass, Jan Bosenberg, Brooks Boutwell, David Irvine, qe, George Sharabidze, Jack Dwyer, and Fredrik!...
(Also thanks to my cats Sage, Dandelion, and Sassafras! Can you spot all 4 of Dandelion's catmeo appearances in this episode?)
Check out the Combo Class Patreon at / comboclass
If you want to mail me anything (such as any clocks/dice/etc. that you'd like to see in the background of Grade -2), here's my private mailbox address (not my home address). If you're going to send anything, please watch this short video first: • You Can Now Mail Me Th...
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[@ComboClass]

In this episode, I explain the unsolved question of the Collatz conjecture, and then show some binary shortcuts and negative extensions to the problem. Thanks for watching! Make sure you’re also tuned in to my @Domotro channel where I put out various bonus content in between these main episodes (and check the description here for other cool links).

[@maiki5962]

Try starting with zero. Infinite loop of splitting zero in half which 0/2 = 0.

[@TymexComputing]

I hate this classes.. but i cant miss it and have to attend it. The pressure of sth falling apart and broking is JUST A HELL FOR AN EGINEERING BRAIN! DONT DO IT IN THE NEXT EPISSEDODE please :)

[@FrankHarwald]

What if current math does have all the tools neccessary to solve the Collatz conjecture but people just hadn't have the right ideas yet?

[@TymexComputing]

@@FrankHarwald Well - i advise you to check out Kurt Gödel's work - his tools including prime numbers :) - it makes a lot of possibilities to think about what else could be possibly correct but unable to prove :)

[@impactrgpdformationrgpd8785]

Very good approach indeed....now let's talk about a Rabbit and a Tortoise trying to prove once and for all that given any length of a racetrack...and any handicap for the rabbit at start of the race...the rabbit ALWAYS ends...catching up with the tortoise ! We can talk about it if you want....thos two rabbit and tortoise have a lot to tell us about how this conjecture behaves ! And its very fun !

[@PaulMiller-mn3me]

You ruined me with different bases… whenever I hear an interesting fact about numbers now, my first response is “that’s only because we arbitrarily count in base 10” !

[@ComboClass]

Some facts are base independent but yeah some other facts are just arbitrarily because of base ten haha. Base six would be better :)

[@DanielTaber-p7f]

Surely that's a baseless conjecture

[@Fictionarious]

@@ComboClass Senary supremacy! Based.

[@Mik1604]

Maybe not totally arbitrary since we have ten fingers.

[@neologicalgamer3437]

​@@Mik1604ancient Egyptians used base 12, the Welsh and Inuit, base 20, and one area uses base 27! There are countless examples

[@MrCheeze]

You mentioned that Collatz is one of a family of similar functions. Surprisingly, those Collatz-like functions show up a lot in research on the Busy Beaver problem - the problem of finding the computer program of a certain number of characters, that computes as many steps as possible before halting (without falling into an infinite loop). It looks like in some sense, the Collatz family may well exhibit the most complexity of ANY mathematical problem that has such a simple description.

[@karlwaugh30]

that's interesting because (& correct me if I'm wrong) but the busy beaver function is often defined in terms of Turing Machines & this binary representation seems to be a lot more fundamental (& less arbitrary than the 3x+1 seems) in terms of binary representation (rather than arithmetic values)... Could you explain more of the "most complexity" and what that really means??

[@MrCheeze]

@@karlwaugh30 Nothing other than the normal busy beaver definition, abbreviated. Collatz-like functions seem to run longer than any other turing machine of equally short definition. We don't know whether Collatz-like programs will always be the champions once we get to longer programs, of course. But if they are, that would imply that just knowing the answer to which Collatz-like programs halt and which don't, would encode a solution to the halting problem (impossible).

[@PhilHibbs]

The negative extension is basically a mirror of “3x-1”.

[@drenz1523]

which is another unsolved question!

[@brosisjk3993]

@@drenz1523 nope, 3x-1 is proven false

[@your_future_self]

Yeah! I played around with the collatz conjecture in tree form in base 6 for a while and I noticed that all of the places where the branches split ended in four. Which I beleive means that any number that could be arrived at in a chain witha 3x+1 operation is exactly 4 more than a multiple of 6

[@ethanbottomley-mason8447]

Well you only get 3x + 1 if x is odd. Odd numbers are numbers which are 1 more than a multiple of 2, i.e. x = 2n + 1 for some number n, then 3x + 1 = 3(2n + 1) + 1 = 6n + 4. So yes, any time you multiply an odd number by 3 and add 1, you do get 4 more than a multiple of 6.

[@m802001]

The guy knows his stuff for sure. I get that the surrounding chaos stands in sharp contrast to the very ordered nature of mathematics, but all that chaos makes me anxious.

[@BillGreenAZ]

Same here...but then again, I have anxiety issues.

[@dbeast03]

Personally, I think the chaos is rather representative of math and science. For as much as we understand, there will always be much, much more that we may never even come close to solving, as even in simple order and concepts such as the Collatz Conjecture, chaos and incalculability is what makes it an oddball problem.

[@Gunbudder]

i love this method of framing, and its how i think about it. what you are really looking for is a chain of numbers that never hits a power of two. a power of two is the only way to get back to one. so really its a question about the nature of powers of two and what triple plus one is really doing. is also interesting that you can construct numbers that have a large amount of repeated divisions in a row, but do not directly lead to one. for example if you square 27 a large number of times, then you can find some interesting paths by working backwards.

[@KaiHenningsen]

27 = 3^3, so squaring that is 3^6, and squaring that is 3^12 ... so 3^((2^n)*3).

[@claytonhiggins7526]

So glad you did a video on this problem, it's one of my favorite unsolved problems! Though personally I think the collatz conjecture is probably undecidable.

[@piercexlr878]

It feels very decideable, but hard to figure out how

[@muskyoxes]

If it's undecidable, then the conjecture is potentially false. If it's potentially false, then anyone could accidentally pick a counterexample and prove it false. I can't decide if this proves it's decidable

[@claytonhiggins7526]

@muskyoxes This actually isn't the case. Since it's possible some sequence could blow up to infinity, you might not be able to prove it's a counterexample because of its infinite nature.

[@drdca8263]

@@muskyoxes Consider the weaker conjecture: “the collatz map has no other cycles other than the (4-2-1) cycle”. If this weaker conjecture is false, then it has a finite refutation. So, if the weak version of it is undecidable, then the weak version is true. So, if the weak version is undecidable, it cannot be proven to be undecidable. That’s not really an issue though.

[@Mihoshika]

"Oh dear. The crazy person in the lab is going into his small tiny yard to talk gibberish again.

[@LAM1895]

It's been about 2 weeks since I saw the Veritasium video about this and I came to the same conclusions about the binary representation recently. It really simplifies the process and shows where the real problem lies: how do we prove that by doing 3x+1 and removing trailing 0s over and over we will always eventually end up with 1? After investigating the problem in many ways I have the feeling that it is the same process as x+1 but the 3 adds a great deal of complexity. It's easy to prove that repeating x+1 and removing zeros will always end up with a 1 -> 2 loop because you're basically just counting natural numbers and any number can be described as a combination of powers of 2 which is the binary representation. So the question would be: can any number be represented with that added 3?

[@ー-ーー]

Here are some axioms I could come up with General: 1. Odd number x Odd number is always results in an Odd number. 2. Adding 1 to an Odd number always makes it an Even number. 3. All Even numbers repeatedly divided by two will eventually turn into a Odd Number. Specific to 3x + 1: 1. In a 3x + 1 sequence the amount Odd numbers will always be less then the amount of Even numbers. 2. There is never two Odd Numbers in a row during a 3x + 1 sequence. 3. As the Even Numbers size increases the amount of divisions by 2 in the sequence increase.

[@lvn5609]

Number 3 is wrong I think. Let's say we have a number that can be factorized as 2q, where q is an arbitrary prime number besides 2. Then, because there are infinite numbers of primes and each next prime number gets bigger and bigger, we can get an arbitrary large even number which can be divided by 2 only once before it becomes odd.

[@gigaprofisi]

I was thinking about this and this is the first video in my feed. Eerie

[@karlwaugh30]

This is fascinating, not least because it seems to move a problem of arithmetic operations behaving kinda chaotically and represents it as a question that seems to sit in symbolic dynamics (not that I have skills in that).... A problem of bits and their overlaps etc. Also there's something nice in the "chop off all the zeros" etc especially as it says 1->1

[@manudude02]

I was thinking about the problem in base 4. All numbers that can be written as a series of 1s (1, 11, 111 etc), will instantly lead to a power of 2 (game over). This will be the same for 2s (2, 22, 222) for obvious reasons. Since we are always looking for higher numbers, we would like to hit a sequence where we cannot hit 111111.... from below. This will be the case for any string of 1s which are not of length 1(mod 3). You would need to prove that you can keep away from a solid string of ones somehow. As a bonus, 100..... is game over, 200.... can never be reached from below (since 133...... is never divisible by 3), 300..... will not be a power of 2, and 00... is zero)

[@skunko1871]

Love this video! Super fun to see binary used here

[@insouciantFox]

This is the Explosions & Fire with math and I'm here for it

[@Ing0s]

These videos are so captivating!

[@lucaspowell8290]

Good shit as always dawg!

[@lugyd1xdone195]

Interesting that if we do this for P-Adic numbers all of those are suddenly possible. Just a question, if we allowed ourselves to count divergent sums as solvable wouldnt it mean that for infinitely large numbers the number only grows, whereas for finitely large numbers if the conjecture is true the number always converges to this cycle? Maybe theres something with divergences-convergences relevant to this problem?

[@Tymon0000]

Problem so simple yet so hard.

[@pulsefel9210]

so doesnt the binary form kinda prove it? it doesnt matter how long the string of 1 and 0 are, tripling plus 1 will always result in a lot of 0s to be snipped and eventually go down. id say you hit the money.

[@mantacid1221]

How is this the first time someone’s told me that Collatz’s first name was Lothar? Like that’s a wizard’s name.

[@lool8421]

i kinda have doubts that it could shoot up to infinity because on average it decreases by x0.75 per operation and to get infinity, you would need some chain that repeats an infinite amount of times however some cycles could exist it's just that multiplication with addition don't go as well as we would like

[@salerio61]

you might have doubts, now prove it :)

[@DeWillpower]

cool to know there are shortcuts in base 3 and others

[@theinternetis7250]

Love these vids, thanks!

[@berzasperdvisiodei6625]

As far as I can tell, negative numbers will fall into a -4 -> -2 -> -1 loop if you alter the rule to be 3x-1 instead of +1.

[@1234567zeek]

I was completely unprepared to be amazed at this unique content! Thank you!

[@pacattack2586]

I don't understand how this isn't proven. I mean doesn't the fact that all the integers from 1-100 do this mean that any integer does this? I mean there's arguement to be made that because two digit even numbers split which way they go that you need to check up to 99, but I'm pretty sure that since all numbers end with an integer from 0 (effectively the 100 case) to 99 you couldn't possibly find a number that wouldn't hit 1 eventually right? Alternatively: Either a number *is* a power of two, or you'll eventually *hit* a power of two ... And the moment you do so it just falls down the pit

[@pacattack2586]

Actually possibly better idea - lets switch number systems and build an INFINATELY LONG NUMBER! Ok - worst case senario is that the last digit is odd. So whatever our infinately long number is it ends in a 1 in binary. Worst case senario again is that when we multiply by 3 and add 1 the *second* to last digit becomes odd, so we want this to end in 01. We will truncate the result until we have an odd digit at the end so when we multiply by 3 and add 1 the new second to last digit becomes odd. this means that we now want our number to be 001 for maximum efficency. so on and so forth and we will eventually build a number that is ...00000001 with infinately many 0s in front. This means that 1 *is* the worst case and it *does* cycle back on itself. QED

[@Kayclau]

If I remember anything from calculus is that going to the extreme always works. So, if the number is 1000000000000...0000001 Then, I'll ad 1000000000000...00000011 And the result will be 11000000000000...0000100 Eliminate the 00 from the end and then ad 11000000000000...000011 That will give me 100100000000000...000100 Continue ->10010000000000...0001 ->110110000000000...0100 ->110110000000000...01 I'm no math expert but, it seems like every number has either a 1; a 2; or a 4 at the end. It truly doesn't matter what number you start with. They all end in a 1 or 10 in binary and if not, the next in the sequence will.

[@grezamisoit]

Tahnk you!

[@SourceEnginePlayer]

the reason it must fall a whole crap ton no matter how much it rises is because of 2 reasons 1: multiplying an odd number by an odd number will always be odd, because if you think about it, odd numbers have an extra piece for another odd number, hence odd+odd is even. When there is an odd number, each of the others get paired up with each other, but there is one more odd, which makes it all in all odd. Adding one will guarantee its even, which will divide by 2. However, since there isn't a guarantee that dividing by 2 will result in an odd, it will always fall more than it rises. Its simple, and because it can never rise more than it falls, there isn't another loop outside of 4-2-1. Its simple??? and 2: what goes up must come down.

[@rogerkearns8094]

If it's Godel unprovable, then it's true. Hope that helps! ;)

[@MuffinsAPlenty]

How do you know this? Did someone prove this recently? It's not obvious to me that this should be true.

[@rogerkearns8094]

@@MuffinsAPlenty I was being somewhat frivolous - perhaps, you are, too, but, just in case you're not, here goes: It's a matter of logic. If a conjecture were unprovable, then there can be no counter-example (as any such would, potentially, provide a contrary proof). If there is no counter-example, then the conjecture is true.

[@MuffinsAPlenty]

@@rogerkearns8094 We have to be careful here. We need _provable_ counterexamples. Maybe you watched the Numberphile video about Gödel's incompleteness theorems, and at the end of the video, the presented (Marcus du Satoy) mentioned that if the Riemann Hypothesis is independent from the axioms, then it is true. And he argued similarly to you: if it's false, then there's a counterexample, and that would constitute a proof. What isn't clear from that explanation is that counterexamples to the Riemann Hypothesis are guaranteed to be provable because we have computable checks for this sort of stuff. As an another example, let me go with the Goldbach conjecture. The Goldbach conjecture states that every even number larger than 2 is the sum of two prime numbers. If the Goldbach conjecture were false, then there would be some even number n which is not the sum of two prime numbers. We could prove this by showing n is even and greater than 2. We could also compute all prime numbers less than n. And we could add all pairs of primes less than n together and see they don't give n. Each of these steps can be done with a finite amount of arithmetic. So a counterexample to Goldbach's conjecture would be _provable_ as a counterexample. For Collatz, it's not so clear to me. If there were a loop other than the 4-2-1 loop, that would be a provable counterexample. You just take a number in the loop and you keep going until you get back to where you started, which would only take a finite number of steps. So if someone shows that Collatz is independent of the Peano axioms, then loops other than 4-2-1 definitely wouldn't exist. But there's another way that Collatz can go wrong: infinite trajectories which blast off to infinity and never loop. The "obvious" way to prove this (just do the process) would never end, so it wouldn't constitute a proof. This is not to say that infinite trajectories are definitely unprovable, but the "just do it" method of proof wouldn't work. So it seems plausible to me that Collatz could be false and unprovable from the Peano axioms. But maybe someone has thought of a clever way to prove that an infinite trajectory exists if you know at least one element in it! If that is the case, then your original claim would be correct.

[@trueriver1950]

Not so fast. If it's Godel unprovable, then you can introduce a new axiom to specify that it's either true or false, without that axiom introducing a contradiction. The "true" axiom has the property you described. The "false" axiom has the effect of introducing one or more new numbers to the set you're considering, thus creating the counter example(s). This is one way to look at the explosion of number types from the naturals to include the negatives, to include fractions, to include the square root of negatives, and do on. For example, the number i is the newly introduced number that's a counter example to the conjecture that there is no square root of a negative number. That conjecture is true if you stick to the reals, but by allowing counter examples we invent the complex numbers.

[@rogerkearns8094]

@@trueriver1950 That's true. Perhaps I should have said, True within the given axiom set. Cheers. :)

[@joshuahudson2170]

The least useful proof would be reporting some number with a different short cycle, but no explanation for how it was found.

[@pierreabbat6157]

It's a hotpo tato.

[@matthewwriter9539]

14:00 Lots of stuff fall down...again... Also...global warming is a huge issue these days.

[@tokajileo5928]

very good but please skip background music next time

[@furnaceheadgames9001]

3n+1 problem

[@user-pr6ed3ri2k]

2x+1 when 0:26

[@user-pr6ed3ri2k]

2x+1 if not multiple of 3 x/3 if

[@ethanbottomley-mason8447]

@@user-pr6ed3ri2k 2 grows without bound with that setup. Notice that if x is a number which is 2 more than a multiple of 3, then x = 3n + 2 for some number n, then 2x+1 = 2(3n+2)+1 = 6n+5 = 3(2n+1)+2, so you get another number which is 2 more than a multiple of 3, so all of these numbers grow without bound.

[@user-pr6ed3ri2k]

@@ethanbottomley-mason8447 yeah I realized this was absurd quicklu after posting this lol

[@phyarth8082]

The Collatz Conjecture is reason why math progress been stagnating. Math lecturer in university introduce The Collatz Conjecture to students with catch phrase that it is imposable to prove problem, for testosterones driven student impossible means "challenge' and all his time student spends to prove that conjecture abandoning other important math problems :)))

[@muskyoxes]

I watched this video and thought "hey, maybe i should scribble a program about this, could be fun". Then i read comments from people doing that for years and thought "okay, not doing that".

[@rajuSaha-eh2ux]

3x+1 hardest unsolved math problem is solved.

[@matthewcarter919]

Math info good. Screaming... not so much.

[@tristanridley1601]

Looking at this in binary is sort of the obvious way to simplify, but most youtube videos seem to ignore it. This feels like it should be such a classic math puzzle. "Oh, convert the base and it's easier. Well, we can consider what the exception would have to look like." If I came across this problem randomly I'd think confidently that I could solve it in a day. And then you look at the number of geniuses who dumped decades into this...

[@trixelpixel5196]

Every time I watch your channel, I’m astonished by how underrated it is. You explain so many things in such a fun way that keeps me interested! Thank you! :)

[@QP9237]

If I told elementary school me that watching a person talk about math/numbers was entertaining/fun, I’m sure I’d call myself a moron… but here I am anyways 😂

[@Asterism_Desmos]

Same, it’s so entertaining

[@sonicwaveinfinitymiddwelle8555]

this is made for kindergarten if you didn't know

[@myownlittlworld9427]

I showed this to a friend, and they said, “that makes sense, but what’s with the clocks?” If you haven’t explained the thing with the clocks to anyone yet, please never do and keep it a secret. Mysteries make the world a more beautiful place.

[@jemmerl]

He's trying to tell us that he has been trapped in an infinite time loop and forced to keep making entertaining educational videos for all eternity. Obviously the conveyed point is that we need to destroy the timeline (breaking the clocks) to free him :)

[@JON-es4wk]

@@jemmerlthe only way to escape is for us to solve 3x+1

[@dickrichard626]

He just likes clocks apperently and there is a video where he uses them to make some points and talks about using different numbers on clocks. Clocks have 12 hours and 12 is an important number to math people, because it is a "highly composite" number and counting in base 12 is arguably better than base 10. The only problem is you basicaly have to relearn math all over again just to be able to think in base 12.

[@josephrissler9847]

The typical way to extend this to the complex plane is to use trig functions. cos²(xπ/2) is 1 for even x, 0 for odd, and sin²(xπ/2) is the other way around. So the collatz function is collats(x) = cos²(xπ/2)*(x/2) + sin²(xπ/2)*(3x+1). This gives you a function which matches the collatz operation on the integers, but can be applied to all complex numbers. The neat thing about this, is you don't *have* to use trig functions for this. You can use any functions with the same integer values. One of my favorite tricks when playing with collatz fractals (using "Xaos", which you can get for free) is to add in sin(xπ)*something. sin(xπ) is 0 at all the integers, so you add literally any function to the mix and get a collatz function that matches at the integers.

[@ethanbottomley-mason8447]

Sin(pi * x) is not 0 at every integer, sin(2pi * x) is.

[@redpug5042]

@@ethanbottomley-mason8447 sin(0) = 0, sin(pi) = 0, sin(2pi) = 0. sin is 0 at every integer multiple of pi.

[@ethanbottomley-mason8447]

@@redpug5042 You are correct.

[@diophantine1598]

I need to try this…

[@s4ad0wpi]

One thing that's worth noting is that if you've tested every number up to N, then you don't need to check any even numbers up to 2N, because they'll be cut in half to a number below N, which you've already checked! You also don't need to check all the Odd numbers between N and 2N, because several of them were probably already a part of patterns you already proved followed the theorem. So that cuts several numbers off your list. It's almost certainly not enough to be used as proof on its own, but probably a good place to start?

[@jejjiz6162]

You would still need to check all odd numbers between N and 2N, otherwise we can inductively say that "since it works for all numbers up to 1, it works for all numbers up to some arbitrary power of 2". That would mean that every number would work. You're correct in saying that even numbers can be skipped, though

[@BridgeBum]

@@jejjiz6162 I think he's trying to say (a) you do not need to check any previously checked number and (b) some odd numbers between N and 2N have been previously checked.

[@KaiHenningsen]

@@BridgeBum True, but you still need some way to keep track of those odd numbers if you want to use that fact.

[@BridgeBum]

@@KaiHenningsen Agreed, however in practice you have to keep track of every number anyway.

[@piercexlr878]

All numbers 1 greater than a multiple of 4 decrease. Same with all numbers 3 greater than a multiple of 8. There's an infinite family of these relationships, but they have few patterns. Since they decrease they either fall to a number thats 4n+3 or they don't and fall infinitely so you can get away with checking 1/4 numbers

[@MarloTheBlueberry]

I've seen Veritasium make a video on this, but I didn't understand it. You made me understand it. Thanks!

[@Fictionarious]

I've spent way too much time thinking about this problem and will undoubtedly spend even more

[@ChannelMath]

it's wasted countless years of good math research. entire careers even. stay. away.

[@Fictionarious]

@@ChannelMath Oh, I know - its just too late for me. I've gazed too deeply into this particular abyss to be saved.

[@leftylizard9085]

I think base 4 would be a useful one to look into. For example, every number that is a sum of consecutively descending powers of four, all the way down to 1, when multiplied by three with one added, becomes a power of four, allowing it to shrink down to one almost instantaneously. It makes sense when written out in base four: 111111111 × 3 = 333333333. 333333333+1 = 1000000000 which, because this is base four, would collpase down to the four two one cycle. This is what happens in binary when a number consists of just 1010101010101 and so on and so forth.

[@biozbran8208]

Amazing video, as always. This playing with numbers always fascinates me and I love that there is a math channel dedicated to this.

[@legendgames128]

Seeing how much simpler the 3x+1 stage is in binary, I feel like the solution likes in cutting the complexity of a number down to 1. (For example, 1101010010101000000000 in binary has a complexity of 13 in decimal, because there are 13 bits before the trailing 0s) And this reminds me of p-adic numbers, specifically, the case for p=2. 13=1101(I will list the complexity in decimal, in parenthesis) 1101(4)->101000(3)->101(3)->10000(1)->1(1) So, maybe this could be useful information?

[@TheGuyvatzian]

I wrote a paper on the Collatz conjecture in my senior year of college. I actually came up with the accelerated version of the formula seen in this video by myself after I read published papers on the problem and a lot of them used a modified version of the formula where instead of 3x+1 if odd they'd do (3x+1)/2 if odd, and I was like "psh. cowards. just get rid of the whole thing!" And after looking at the problem that way I realized that the big question in the conjecture is really whether a power of 2 will eventually be reached. This led me down a rabbit hole where I realized that the difficulty in the problem comes in the unpredictable way that the 3x+1 step messed with the prime factorization of x. But while that took me a while to realize, it's really not that much progress. To find a pattern in the Collatz conjecture is to find a pattern in prime factorizations

[@legendgames128]

@@TheGuyvatzian The +1 really messes with things. Multiplying by 3 just adds another 3 to the factorization, but as soon as you add 1, everything goes haywire. What would happen if we tested for big prime numbers? I wonder if that could reveal anything.

[@davidhand9721]

This is the path I've been thinking about, too. However, I don't see the increment as being the difficult part, but the multiplying by 3. Adding 1 to a binary number is very simple: eliminate all the trailing 1 bits, add a 1 bit at the first zero. Multiplying by three, i.e. adding x and x

[@AnthonyFlack]

​@@TheGuyvatzian watching this video, I was also thinking that you may as well make the odd number process be (3x+1)/2, because 3x+1 will always turn an odd number even, so the next step will always be to divide by 2. This makes it feel intuitive that the series will always trend downwards over an arbitrarily long time, since there are the same number of even numbers as odd, and even numbers reduce the total by more than odd numbers add to it, proportionately. So in order to keep going up forever, the sequence would have to land on more odd numbers than even, all the way up. And half those even numbers are doubly even, and half of those are triply even... and of course if you ever land on a power of 2 it's game over. Is there a streak somewhere that favours odd numbers all the way up? Again it seems intuitive that this couldn't work, but intuition isn't proof. My suspicion is that in order to work, a sequence would have to favour odd numbers to the degree that it trends towards landing on only odd numbers as it gets larger. If a sequence ever falls back on itself you have a loop, so if the sequence ever lands on the same number twice, you've won a hundred million yen. But again, it seems "intuitive" to me that you could never reverse a 3x+1 by applying x/2 or more 3x+1, except when x=1.

[@AnthonyFlack]

@@davidhand9721 - I was thinking along similar lines too; if you could show that the operation always trends towards reducing digits, or adding more zeroes than ones... but of course it can't be that simple. This is just how you get hooked into it.

[@Snidbert]

Why do pop math channels always call it the “3x+1” conjecture? Shouldn’t it be the 3n+1 conjecture, since it deals with natural numbers rather than reals?

[@skeksis268]

I’ve been working on that binary representation of the collatz conjecture on and off for a few years. It’s quite fascinating and has some nice properties, I should write it up…

[@zrodger2296]

Please write it up, and congratulations if you do. I've found some interesting problems over the last few years and have exhaustively solved them on my computer (cool stuff, but nothing really important) but never really finished them, nor have I written up the results. Well, a new year is coming up...

[@ionmurgu783]

was a good idea. Unfortunately need to understand "the scope" of demonstration is to show , not leaf effect, not root inversion, which and where I don't think will help. Proof in Mathematics mean #Dead_Of_Science See #FermatLastTheorem!

[@epistax4]

Thanks for making this. A few years ago i found the same thing about treating the problem in binary. It's funny that multiplying by 3 is so easy in binary. You know you've converged as soon as you get a 10101.. pattern. Dropping the trailing 0's super satisfying too. It just doesn't get us any closer. If you want to make things harder on yourself instead, try 3x+1 in prime factors. Dropping to the odd is just as easy as with binary. The 3x is also trivial. But effect of a +1 on a prime factored number is extremely arbitrary. Obviously (or not) every single factor is changed by the +1, in a unpredictable fashion. I've tried to find some sort of physical phenomenon that is model by 3x+1 but nothing.

[@swiddle1]

I'd like to see how this looks in base three and in base six.

[@popularmisconception1]

with negative numbers, you're basically playing a 3x-1 version of the game with minus sign as a decorator. but it's a nice example that it might make sense to try out modifications to understand that more cycles can in fact emerge (and how and why). maybe a 5x+1 version where all numbers divisible by 2 or 3 would be ridden of these factors, or 5x+3, or 7x+1 removing 5s, 3s and 2s from prime factorization... btw, prime factorization is also a good way how to think of a number, but adding 1 is tricky with this representation. or is it?

[@deemaske9143]

honnestly, i love how your decorative stuff keeps falling or breaking

[@ratlinggull2223]

I've been making a small game about the Conjecture and I actually use binary as a strategy to play it. Nice to see this video popping up now.

[@VinyJones2]

What if someone prove that it is improvable

[@electronlibre7921]

Collatz is also an amazing problem to take a look at as a grammar problem and not a numerical problem. That leads to interesting behaviour of paterns. Where you can really pin point a Rabbit...trying to catch a Tortoise !

[@SirRebrl]

Paused shortly into the video, already having thoughts about what was going to make binary interesting for the conjecture. Decided to pick a random binary number and run through the steps. "Chop off all the zeros" was obvious. I didn't think of your 3x+1 strategy, but I was iterating on a whiteboard and focused on preserving space for more iterations. I did come up with a simple system of counting the 1's belonging to each position, then translating evens to 0 and odds to 1. My partner got home as I was doing that and immediately called me a nerd, so I nerded up and explained what I was doing, including the trick for dividing by 2, comparing it to dividing by 10 in decimal. Was entertained to listen to you run down everything I had just said to her when I resumed the video. =D (My random number ended up being 1625 btw, and I got through 20 iterations, reaching 175)

[@leavealoner]

Is this easier to solve by looking at it the other way around? start with the loop, and apply the reverse of both equations to every single number in the number line if it is applicable. like, all numbers can be doubled to see which even number they were derived from, and all numbers can be minus-oned to see if the result is divisible by three. isn't it easier to see wether or not that fills the number line or not?

[@LAM1895]

I tried that for fun. I wrote a Python script that creates a binary tree of all possible paths starting from 1. Here is the heart of the code: # Calculate left child value if (value - 1) % 3 == 0 and 3 < current_depth < depth: left_child_value = (value - 1)//3 if left_child_value % 2 != 0: tree.add_edge(value, left_child_value) queue.append((left_child_value, current_depth + 1)) # Calculate right child value right_child_value = value * 2 if current_depth < depth and right_child_value < 10000: tree.add_edge(value, right_child_value) queue.append((right_child_value, current_depth + 1)) I added a limitation to the value of 10 000 because since we're dealing with powers of 2 the numbers and the spread of the tree quickly become huge so it takes a long time to calculate and render. I came to 10 000 by taking the biggest number in the sequence of 27 and rounding up because that particular sequence is an interesting outlier. If you think about it, 27 is on the same depth as 2^111 which is 34 digits long. If you want to do the same I recommend you render the image in SVG format to eliminate the resolution problem.

[@Gunbudder]

don't you dare Collatz me! i've already spent way more time than anyone ever should studying this dang thing. its a black hole of mental energy!

[@peterlaloli6279]

Fantastic video, your content is an absolute joy to watch.

[@aaronhorak710]

1 mod 4 is always followed by a smaller odd, 3 mod 4 by a larger odd. Easy to see in binary, the former ends in 01 the later in 11. Been working on properties of assumed cycles and trying proof by contradiction. Enjoyable presentation!

[@PW_Thorn]

The first time I heard about 3x+1, I immediatly thought that a proof of it can be achieved by seeing parterns and relationships and properties for integers in base 2 and 3. And that's here the first YT video going quite that way 👍

[@Tyshun-r6c]

your rule can be represented by the following equation: \frac{x + 1}{2} & \text{if } x \text{ is even} \\ x + 1 & \text{if } x \text{ is odd} \end{cases} This equation captures the essence of your rule: adding 1 to the input and, if the result is positive and even, dividing by 2; otherwise, leaving it unchanged

[@QP9237]

Can you do a video on hypercomplex numbers or the hopf fibration? Always enjoy your videos, so would like to see you tackle these topics!

[@ComboClass]

Thanks! There will be an episode about hypercomplex numbers at some point. Not sure about a hopf fibration episode but it’s certainly possible :)

[@QP9237]

Awesome man looking forward to it 😁

[@josueantovani8019]

i think that if you use fractions instead of natural numbers, it would be proven to be wrong in a second lol

[@Guywiththetypewriter]

It saddens me you did this as I fear you have got so close to a proof here. I was told about your reasoning when I contextualised it a bit more broadly. "The 4,2 1 pattern occurs the moment you get to a hamming weight of 1(a single 1 bit) no matter how large the number is" And multiplying by 3 only ever shifts the addition to be done 1 bit to the left... But get this... When you add one to a multiple of 3, either the hamming weight stays the same or will remove n-1 from the hamming weight. Where n is the size of the least significant run of ones... Add to that the fact that, whilst not proven just yet (but man's close) that the only time hamming weight increases is when the 1 bit is added to an existing pair (hence being negated by the eventually termination due to the +1 step) the hamming weight of the binary always trends downwards.

[@44Mat]

Couple months back I asked about the relationship between this equation/function and its plotted graph/points vs the inverse of the equation if it also gives an inverted mirrored graph/points....And what do we call such an equation if the inverse doesn't give opposite answers or plots Replace f(x) by y Solve for y.....I wonder if an A.I hallucination could go with me through this rabbit hole and we loosen up a little and refuse to be rigid when trying to solve this one

[@ChannelMath]

here's a better plan of attack, my young padowans: 1) learn how independence proofs work 2) learn forcing 3) prove the Collatz conjecture cannot be proved from the standard axioms of math (see Hilbert's 10th problem for inspiration)

[@imeprezime1285]

Why would anyone offer lot of money for solution to this problem? There are 1000s of similar unsolved problems. Compared to some serious unsolved conjectures in math, importance of this one converges to 0

[@suan22]

I think we "just" have to prove that if you start this procedure with any positive integer number then it will eventually reach a number in the sequence that is less than the starting number. Then induction would prove that this means every number will eventually reach 1, no?

[@otonanoC]

Four hours ago this was posted? Lets get it on.

[@tjdrennan7147]

Just change 1 + 1 to equal 3, the unity of 2 creates something new a third, do this and 3x + 1 becomes infinite

[@ephemera2]

I am nearly 100% certain that I have at the very least broken new ground on this conjecture

[@emilyrln]

I can't wait to learn the name of your short-haired black kitty!! (Don't tell me; I want to find it in a video 😂)

[@aewfawef963]

people that are into math are so, so, weird. yes, I mean you. yes, this is a compliment.

[@hancocki]

My world's on fire. How about yours? Oh. well... Hey now, you're an all-star.

[@wryanihad]

Can we prove collatz conjecture by induction method?

[@Joe_Payne]

Negatives are basically positive except its 3x-1 instead of 3x+1

[@triffinne]

I saw the globe on fire and immediately thought "Oh no, a model of global warming" 😆

[@gm2407]

What about factors as well? Will that aid in demonstration of this?

[@815TypeSirius]

It reaches the same loop as one because you add 1. Enjoy the yendolars.

[@alejrandom6592]

I dropped my phone and then the whiteboard fell. I'm sorry.

[@barakap2372]

Wow I really like your videos. Look up to you.

[@alejrandom6592]

Heeey aren't you "The Guy" from spy kids?

[@DanF8]

the globe's on fire 🌍🔫

[@FrankAnzalone]

Shouldn't the negative version be 3X −1?

[@yakuzzi35]

I love you maths Jack Harlow / Mr Tumnus

[@splodeyferret]

I don't want to set the world on fire~

[@Tletna]

Oh no, has Domotro been sucked into the neverending story of 3x + 1? It is an interesting conjecture.

[@ComboClass]

Haha. I've been planning this episode for a while (I brainstorm episodes way in advance) and it will probably be my only Collatz-related one for a while (apart from maybe some more bonus content about it on my Domotro channel) but we might return to the topic someday :)

[@Meanslicer43]

so many broken clocks.

[@rrock2025]

Why the ellipsis?

[@TimJSwan]

There is a collatz fractal, but you showed the mandelbrot set fractal instead. 🤦🏼‍♂️

[@MuffinsAPlenty]

You might want to take a second look at your fractals.

[@ComboClass]

No, the fractal I showed was not the Mandelbrot set. And there’s not “a” collatz fractal, there are various fractals that could be made out of extensions to it. The one I showed is linked in the description if you want more details

[@Very.Crazy.Math.Pistols]

I'm not an expert in diophantine equations at the moment, but if a number falls into one of the integer solutions that stand in for x, we're there: solve diophantine 3x + 1 = 2^y.

[@Very.Crazy.Math.Pistols]

Obviously we need to prove that the algorithm / 2 , 3x + 1, will always end up on one of the solutions x, of ( x,y ) 😅 👹

[@BillGreenAZ]

Domotro: "Computers checking a bunch of numbers isn't going to prove something about the infinite amount of natural numbers..." I believe there was one conjecture that fell into this situation of being proven to a large number but was later proven to not necessarily be true. I can't remember the name of the conjecture, but this principle supports your statement.

[@ComboClass]

Yeah there’s a few problems like that, and sometime I might even make a future episode topic about conjectures/patterns that break at large counterexamples. A great written work about a similar thing is “the strong law of small numbers” by Richard Guy.

[@BillGreenAZ]

@@ComboClass That would be great. Thanks Domotro!

[@ethanbottomley-mason8447]

Look up Skewes's number, it is a good example of things failing only at very large numbers.

[@BillGreenAZ]

@@ethanbottomley-mason8447 Thanks for the reference.

[@cjrm15macpherson20]

you said a whole number which includes 0. 0 is even and if you divide it by 2 you get 0

[@ComboClass]

The term “whole number” is actually ambiguous and used differently by different sources. I also clarified that it had to be bigger than 0 for the normal process and showed that 0 loop later