UNCRACKABLE? The Collatz Conjecture - Numberphile

Catch David on the Numberphile podcast: • A Proof in the Drawer ...
Professor David Eisenbud on the infamous Collatz Conjecture, a simple problem that mathematicians may not be "ready" to crack.
More links & stuff in full description below ↓↓↓
Extra footage from this interview: • Collatz Conjecture (ex...
Prof Eisenbud's 17-gon: • The Amazing Heptadecag...
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[@numberphile]

Catch David on the Numberphile podcast: kzbin.info/www/bejne/b6qUc3qso7msh6M

[@raydarable]

Why does the book show 1 going back to 2? Shouldn't it go to 4?

[@ashkanledu2282]

​@@raydarable Cause as he explained in the video, an odd number multiplied by 3+1 is always even so he goes 1 step further and divides it by 2 So the formula would be : if it's even : n/2 if it's odd (3n+1)/2

[@ashkanledu2282]

@@tejaspathak375 Check my comment below, I get a bit deeper into this question

[@raydarable]

@@ashkanledu2282 Thanks, didn't realize the cover was skipping that step.

[@foxleo6729]

Isnt this just a fractal if you layer all these solutions?

[@scottanderson8167]

“Seven seems to be an odd number.” That’s why he’s a maths professor.

[@user-sx2lc4jt1r]

What d u mean

[@HN-kr1nf]

@@user-sx2lc4jt1r he is clearly extremely skilled and educated in the field of mathematics

[@debashisbarik4805]

😆 😆

[@sohee7597]

>_>

[@powerplay.556]

He's certainly no English professor. Love when people say "prehaps" (0:08) and then think they're smart.

[@simonreinsperger718]

"Wow... 16, very even number!"

[@sarahvan3826]

Because it's a power of 2!

[@nohaylamujer]

That's right: it's quadruple even.

[@arunpathak9851]

Shut up

[@thatoneguy8966]

Square root, fourth root, divide by two, divide by four

[@ibite100]

Your problem??

[@jalepenopvp2300]

This guy has legit the most calming and comforting voice I’ve ever heard

[@Nerine98]

He reminds me of my university professor so it actually makes me nervous as it reminds me about the work I have to do damn

[@Asofia20]

it’s like mathematics with Bob Ross

[@-SpaceWizard-]

So frickin accurate 😤@@Asofia20

[@badhombre4942]

This must be the formula that banks use, to calculate fees, that reduce my account to 1.

[@mitodrumisra8972]

🤣🤣🤣🤣

[@mr.ditkovich6379]

🤣

[@Eleni_E]

I'm reminded of the wikipedia phenomenon where, if you click the first link in almost any article that isn't in parenthesis or italics, and keep doing that long enough, you will eventually end up at philosophy.

[@traso56]

tried nuclear power plant ended in philosophy tried C language, i'm looping when i reach math they have read this :o

[@typo691]

And your comment reminds me of the six degrees of separation.

[@DeathBringer769]

The Wikipedia degrees of separation from Philosophy... it usually gets you there within 5-10 clicks from my experience, lol.

[@1jerrycamacho497]

That's because philosophy is the source. Without philosophy, we could not be as advanced as we are. This is common knowledge to philosophers.

[@I.amthatrealJuan]

Also try that with Hitler

[@Croxmata]

And as we learned, Brady was most likely to choose 7 out of all numbers from 1 to 10.

[@SatanicBunny666]

Actually, some studies have been done on this and turns out, if you ask people to pick a number between 1 and ten, 7 and 3 are the most common answer.

[@Halo3Matalix]

But if there is a study showing that people pick 7 or 3 as a common answer, doing the test again would result in different numbers or the same numbers. we don't know if the data is correct simply because we can't tell if people are really just randomly picking those numbers or picking them on purpose because the information they got from the initial test. i guess what i'm trying to say is the moment we "record" something is the moment that bit of information becomes irrelevant.

[@MugenJinSan]

There's actually a video about this on this channel, that's why Vexatos said that.

[@furbyfubar]

Also, he obviously *knew* the subject he was asking about, so he was not very likely to pick 1, 2, 4 or 8 given how short a chain that would be... 5 would also be sort of short. It just made for the best example here.

[@MadTiger20001]

People often pick 7, sometimes 2 or 3. People like primes and 5 is too central, it seems too obvious.

[@finchisneat]

My friend in high school told me about this nearly 20 years ago, so cool to see a video about it!

[@megauser8512]

I notice that if you start with 0, you don't get to 1, but you keep looping back to 0 over and over again, since 0 is even. Also, at the end when they ask "Why not 3*n - 1?", I thought "Hey, if we do the 3*n + 1 problem with negative numbers, then if we let n = -m, where m is a positive odd number, then we get 3n + 1 = 3*(-m) + 1 = -3*m - (-1) = - (3*m - 1), and if m is even, then we get 2*n = 2*(-m) = -(2*m), so the negative cases reduce to the 3*n - 1 problem for positive n (if we take the absolute value)!"

[@JohnSmith-xb4ux]

It's amazing how many teachers shouldn't be teachers , after just paying a few minutes of attention while I was eating I just posted the solution in the comments section.

[@asiamies9153]

@@megauser8512 0 not natural

[@sahiba2297]

@@megauser8512 The Conjecture is defined for Positive Integers ≥ 2

[@yashrawat9409]

*" Don't Judge a b̶o̶o̶k̶ problem by its c̶o̶v̶e̶r̶ statement "* *- Collatz Conjecture*

[@robinaylott1264]

Genius!

[@JimCullen]

Man I chose 4 initially. Worst possible number…

[@blasttrash]

i chose 1. :P

[@AlexKing-tg9hl]

Jim Cullen 2

[@thefamousarthur]

4 goes to 2. 2 goes to 1.

[@roylavecchia1436]

I chose 0.

[@WordoftheElderGods]

I picked 8.

[@andlabs]

My first introduction to the problem was one of the example in former Bell Labs Unix co-pioneer Jon Bentley's book "Programming Pearls", which is a collection of columns on the theory and practice of software design he wrote for seminal comp sci journal Communications of the ACM over the years, expanded and with exercises. One of the exercises in one of the early chapters was this conjecture (in the Second Edition it's Column 4 Question 5). In the back of the book is a collection of hints. Here is the hint for this question. It has stuck with me ever since I first read it: "If you solve this problem, run to the nearest mathematics department and ask for a Ph.D."

[@elrak0219]

Is there any reason that “n” can’t be negative? if you do it with a negative variable, it just ends up as -1 and not 1.

[@danishqureshi8583]

elrak0 2 I think it's because of the "+1", adding a positive to a positive is different than adding a positive to a negative, since 5+1=6 but -5+1=-4. But I'm no math genius, so idk. For example: -5 -5(3)+1=-14 -14/2=-7 -7(3)+1=-20 -20/2=-10 -10/2=-5 Cycle repeats.

[@elrak0219]

Danish Qureshi That’s exactly how it works, I believe there are four total loops fot starting at a negative(if you do a quick google search for 3n+1, you should find a Wikipedia page that has more than enough information on variation, loops, mathematics, etc.)

[@alexandertownsend3291]

@@elrak0219 n could be negative, but then it wouldn't be the Collatz Conjecture. A conjecture is an educated guess in math. The collatz conjecture was an educated guess by the mathematician Lothar Collatz. The conjecture is something like, "I think that every positive whole number n goes to one under this process". While these are probably not his exact words you get the idea. He said nothing about n being negative. You could ask about negative numbers, but that is its own separate, but still interesting problem. I hope that clears things up.

[@tommyencrapera1629]

What is the formula trying to decipher? Like why did they choose those numbers 3x+1 and not 4x+1 or 3x-1 and so on , what is the end game ? Is it the Only formula that always falls back to 1 and they dont know why? I'm trying to figure out what it would prove if there is a number that goes onto infinity or completely forms a separate loop , then what would that determine? Sorry for not understanding and seeing what I'm missing .

[@nimets123]

it is interesting to look at the problem with binary numbers. dividing by 2 is just binary shifting to the right, because lowest bit is 0 for even number. 3n+1 is shifting the initial number to the left, adding the original number and then add 1. the middle path is reached when the highest bit is 1 and the rest is 0. they then can be reduced to "1" by shifting to the right.

[@esotericVideos]

The first part of every numberphile video gets me excited to try to solve a problem, the second part convinces me that there's no point in trying because tons of people have already tried. :/

[@falchion3568]

choose a number from 1-10 "42"

[@marcospinto9380]

3.3333333

[@sohee7597]

9.085574564317754211 , yeah, that one in particular

[@sohee7597]

9.085574564317754211 , yeah, that one in particular

[@arpitdas4263]

Wonderful

[@larho9031]

42 -> 21 -> 64 -> 32 -> 16 -> 8 -> 4 -> 2 -> 1

[@numberboxgamer]

This is literally just "4 is cosmic" in math form lol I appreciate that very much.

[@kevingil1817]

0:16 He chose seven on purpose! He knows it's the least random number! Bold strategy.

[@blu4able360]

Thank you for making this video!

[@numberphile]

+Jake S (blu4) thank you for watching it. Please show your friends. :)

[@eugenebayak6807]

+Numberphile Ok (no)

[@rafa3lico]

wat

[@CoolJoe330]

Nissan > Suzuki

[@1951split]

+Numberphile Can you explain to me why sequences that follow any number that solely consists of a lot of ones (like 1111111111111111111111111111111111111111111111111111111111111111111111 ) always follow the next general rule: {meaningless number}25{a lot of zeroes}4 {meaningless number}25{a lot of zeroes}2 {meaningless number}25{a lot of zeroes}1 {meaningless number}75{a lot of zeroes}4 {meaningless number}75{a lot of zeroes}2 {meaningless number}75{a lot of zeroes}1 and so on... the {meaningless number} gets bigger and the number of zeroes gets lower, until there are no more zeroes left and then the sequence turns ""normal"" like any other sequence...

[@ljfaag]

I found a counterexample, but the comment section is too small to contain it :P

[@justlikedirt9634]

Then your lying, or else you would take every way possible to show everyone collaz is wrong. Stop faking smarts to get attention.

[@IceMetalPunk]

+JustLikeDirt I think you missed Ijfa's reference...

[@sworupadhikari7319]

+JustLikeDirt. Its a joke. Fermat wrote that he had a an elegant solution to his last theorem but the margin in the book he was studying was too small to contain it.

[@rohitg1529]

+JustLikeDirt it's a joke. it's what Fermat did about his theorem

[@philofblood855]

made my day

[@MiaVilleneuve]

When he said any fourth grader could understand it, I was like, that's probably an exaggeration. Then he explained it and sure enough I remembered reading about it in a children's math book called "Math For Smarty Pants" in elementary school.

[@Xonatron]

Ha, awesome.

[@maki6203]

today was my first day at uni and my professor mentioned this conjecture. i was intrigued so i decided to look into it more THANKS NUMBERPHILE FOR COVERING IT

[@Faisal710]

I am working on Prime numbers and I am amazed to see your expression they are extremely useful in my work Thank you man

[@simonmultiverse6349]

The cover of the book is taking a short cut. It has an arrow going from 1 to 2 so it's showing two steps n-> (3n+1)/2 (combining the multiplication and the division by 2).

[@air9music]

"I bet the person who found it thought maybe he was on to something..." I'm willing to bet the guy who found the tree for 63,728,127 wasn't sitting down and doing it 😂

[@CineGeeks001]

Hold my java program for finding the steps

[@davidhahnert2914]

I didn't know that mathematics had a Bob Ross. Homeboy mentions trees, clouds, and visual patterns in his explanation.

[@jess4750]

I love watching your videos! They always give me something new to think about, as a future mathematician it's great to get a glimpse of what I could be working on!

[@travispetit2410]

Finally!! A video about the 3x+1 problem :) Liked

[@apollion888]

Great speaking voice, I'm sure you are a terrific teacher

[@haoyu7937]

I have a truely marvelous proof to the collaz conjecture whitch this planet is too small to contain

[@user-lj8mr6fk6s]

)))))actually, math is going to end in this way for every single new proof someday)))))))

[@arpitdas4263]

Outstanding move

[@bambolere]

Fermat is everywhere

[@air9music]

Is that you, Fermat?

[@alexandertownsend3291]

Aaahhh! Zombie Fermat!

[@davidwiatr8402]

Thanks for the book reference. I ordered it!

[@MrJeevies]

I love this channel :)

[@billlarrison3251]

Fascinating, you have made me understand. Thank you.

[@dragoda]

Veritasium also made a great video about this conjuncture. I love math and your channel. Please never change.

[@TimJSwan]

You did it!! The Collatz! Thank you! Now, I am about to be away from electronics for 9 months. At least I'll know that you guys made a video on the conjecture like I asked. I tried to prove it as an exercise. Going to watch the extra footage, of course. :)

[@garrett3883]

I was messing around on my calculator and I think I found a similar problem. It has 3 rules. If even dived by 2, If divisible by 3 dived by 3, IF the number isn't divisible by 2 or 3 the multiply by 5 and add 1. do this and It always seems to get stuck at the loop 6, 3, 1, 6, 3, 1. Or depending on If you divided by 3 or 2 first 6 ,2, 1, 6, 2, 1. This works regardless of the number.

[@TimStamper89]

The Garrett guesstimate ?

[@yashiAxen39]

I love this channel. This gives me some space vibes!

[@moudar981]

Very nice video Thanks Prof. Eisenbud

[@doodlefox9837]

I like how it started like the fibonacci sequence; 1, 1, 2, 3, 5. I wonder if it follows that pattern even longer with a bigger tree..

[@logan2669]

I actually bet my friend 20$ that he couldn't do the 3n + 1 without getting to one I won the bet

[@cheesywiz9443]

xD

[@poonamroy5873]

Excellent video!!

[@evanherk]

what a lovely teacher.

[@reziik6904]

I just learned to make simple stuff in python yesterday and my second program was making a loop to take a random number between a set range and do this.

[@Albeit_Jordan]

I think the key to this conjecture takes the form of yet another such - if we find the probability of *n* either being odd or even in the simple expression *x∙y=n* (where x and y are odd and even whole numbers) then I reckon it'd just be a game of permutations from there.

[@theroadie7537]

I have been running this project on BOINC for so long i forget when (2 years?) and now i know what it is doing.

[@yolandafraga5182]

Love this videooo

[@BurakBagdatli]

Thanks for ruining my life. I've been absolutely obsessed with this since the video came out and cannot function as a productive member of the society anymore.

[@manijoy6437]

i feel your anguish

[@tgrizz213t]

so does he still find a pattern?

[@thesheq5023]

Burak Bağdatlı zero will not result in one

[@zer0python]

How far did you get? I went as far as translating the function over 4x. But stopped working on it. I'm sure if I analyzed the x sequences, I'd end up just as lost as everyone else. :-) -- I prefer spending my time on the factoring problem though. It's just such a fun challenge. For anyone else who wants to play with it (yanked from my notes): Lets recap, originally we defined the function as follows: f(2n) -> n/2 f(2n+1) -> 3n+1 but with our new found knowledge we can define it further as: f(4n+0) -> n/2 f(4n+1) -> 4 * (3 * (n/4) + 1) f(4n+2) -> 4 * floor(n/8) + (2 + -1^((n/4)+1)) f(4n+3) -> 4 * (3 * (n/4) + 2) + 2

[@esotericVideos]

Don't solve it before me! lol I've been working hard on it. I keep thinking I have a thread on a solution.

[@stephen2876]

73 takes 73 steps to reach 1

[@numberphile]

+Stephen H I've not checked but of so, that's awesome. How many numbers have that property?

[@rafa3lico]

so many questions can be made.. damn

[@stephen2876]

My bad. Some poorly written code on my part, I forgot to set up the count properly (oops). 73 actually has 115 steps. The only number in the first 100 numbers is 5, although the number 91 comes close with 92 steps.

[@stephenkamenar]

73 takes 73 steps if you use the shortcut he mentioned to do (n*3+1)/2 as one step

[@BradenBest]

For n = {1 ... 9999}, the highest number ever reached is 27,114,424, which happens at n = 9663, step #48 of an astounding 183 steps. ===== N = 9663 ===== 9663 28990 14495 43486 21743 65230 32615 97846 48923 146770 73385 220156 110078 55039 165118 82559 247678 123839 371518 185759 557278 278639 835918 417959 1253878 626939 1880818 940409 2821228 1410614 705307 2115922 1057961 3173884 1586942 793471 2380414 1190207 3570622 1785311 5355934 2677967 8033902 4016951 12050854 6025427 18076282 9038141 27114424 13557212 6778606 3389303 10167910 5083955 15251866 7625933 22877800 11438900 5719450 2859725 8579176 4289588 2144794 1072397 3217192 1608596 804298 402149 1206448 603224 301612 150806 75403 226210 113105 339316 169658 84829 254488 127244 63622 31811 95434 47717 143152 71576 35788 17894 8947 26842 13421 40264 20132 10066 5033 15100 7550 3775 11326 5663 16990 8495 25486 12743 38230 19115 57346 28673 86020 43010 21505 64516 32258 16129 48388 24194 12097 36292 18146 9073 27220 13610 6805 20416 10208 5104 2552 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 Highest: 27114424

[@seadub4944]

2^950 Now i just need a universe large enough. Maybe ill find a lever large enough there too. 🎣🎣🎣

[@parsec5366]

You can: 2^950 = 9516908214257811601907599988159363584840065290620124537956939899622020205826587990689077212775400643774711832257235027522909345571487396529861315719055325605011013378863743193233193022939505515969530853007049198118833591724018432564205433218231411731277088674906521042072098232413978624 Steps 950

[@mobidick6064]

Sir, u explain very well

[@BucketCapacity]

I have studied the Collatz Conjecture in my spare time. The closest thing I got to anything useful was, when n =/= 24*2^x+1, x in the integers, n going to 1 in the collatz function is equivalent to there existing integers a and b such that 4a | n+b and b | n +4a. This does work for certain n in the form of 24*2^x+1, but not all of them (ex n = 193)

[@juliusalbe2070]

So every number is eventually gonna hit a power of 2 and is done than. Even if they tend towards infinity, they would allways somewhere hit a power of 2 and than go all the way down to 1.

[@numberphile]

But what if it gets caught in its own loop BEFORE it hits a power of 2 and drops to the ground --- for example see the extra footage linked in the description... If you use 3n-1 instead of 3n+1 that happens to 7....

[@buildasnowman4601]

Why would they always hit a power of two?

[@austinmitchell8846]

Maybe. We assume it is true, and all our evidence so far points to it being true, but we can't prove that it is.

[@dariusdurian1910]

This is mathematics that involves powers of 2. Something math could do but not us humans

[@YourMJK]

+BuildA Snowman Because the only way for a number to become smaller (to become 1) is by dividing by 2, so if all will eventually get to 1 they must hit a power of 2

[@culodesobra]

it would be interesting to analyze the lenght of the branches

[@enlongchiou]

Reciprocal of (3m+1)/2^n rule of Collatz conjecture is (2^n*m-1)/3, can use (2^n*m-1)/3=m for both, it's only solution is m=1.

[@Transyst]

This only shows that it doesn't return to the same number after just one 3m+1 step, no matter how many consecutive /2 steps afterwards, except for m=1. But it doesn't exclude cycles with multiple 3m+1 steps, or the possibility that it doesn't end.

[@HL-iw1du]

Please post more videos Enlong

[@TheHereticAnthem20]

5:43 and then eVENTUALLY

[@accounttest1660]

eV *ENT* U A ll y

[@morgiewthelord8648]

Your point?

[@leobitencourt4719]

I think it has to do with an odd number multiplied by 3 adding 1 always being an even number, but an even number divided by two could give ya an odd number or another even number, so it kinda eventually forces it into powers of 2 and we all know it's downhill from there

[@kb9880]

Yeah that's the intuition behind the "conjecture" but we can't prove that eventually it will reach powers of two.

[@thej3799]

@@kb9880 I don't think you can ever prove by doing it the hard way from here the upper limit of a diverging infinite set of " " prime values when he talks about records he's making a very important point because what is the biggest number in the list of records how can you define that as being the biggest without a sense of something greater it means nothing without the idea that there is yet another number. I was thinking this morning about this powers of two thing it's weird to see it show up in my KZbin feed because I wasn't saying much of it out loud. I was thinking of the difference between square roots cube roots and different numerals versus decimal integers so I found the whole thing fascinating to think about but part of it was because you're dealing with recursion once you start adding a decimal and going beyond and so you're back into these recursive trees when thinking about decimal roots it's weird that this would show up I like this video.

[@WaitTillNextYearChi]

The Collatz Conjecture states that for any positive integer n, repeated application of the Collatz function f(n) = n/2 (if n is even) or f(n) = 3n+1 (if n is odd) eventually produces the number 1. To prove this, we will first show that f is a well-defined function from the positive integers N to the eventual outputs {1,1,1,...} of the Collatz function. We will then show that f is injective, meaning that different inputs produce different outputs, and that f is surjective, meaning that every output is produced by some input. This will establish that f is a bijection and hence that the sets N and {1,1,1,...} have equal cardinality, proving the Collatz Conjecture. To show that f is well-defined, suppose that we have two different sequences of Collatz function outputs that start from the same input n and eventually reach different outputs. Then, by definition of the Collatz function, these sequences must differ at some point. However, this contradicts the fact that the Collatz function is deterministic - given an input, there is only one possible output. Therefore, the Collatz function is well-defined. To show that f is injective, suppose that there exist two different inputs n and m such that f^k(n) = f^k(m) for some k ≥ 0, where f^k denotes the k-th iterate of the function f. We will show that this leads to a contradiction. Suppose that n and m have the same parity (i.e. they are both odd or both even). Then, applying the Collatz function once to both inputs produces two new inputs, either both even or both odd, which are still different from each other. Since the parity of the inputs remains the same, this process can be repeated indefinitely, producing an infinite sequence of different inputs. Therefore, if n and m have the same parity, they cannot produce the same output after repeated application of the Collatz function. Now suppose that n and m have different parity. Then, without loss of generality, suppose that n is even and m is odd. Then f(n) is even and f(m) is odd, so f^k(n) and f^k(m) will always have different parity for any k ≥ 0. Therefore, if n and m have different parity, they cannot produce the same output after repeated application of the Collatz function. In either case, we have reached a contradiction. Therefore, if n ≠ m, then f^i(n) ≠ f^i(m) for all i ≥ 0, and hence f is injective. To show that f is surjective, fix k ≥ 1. We will explicitly construct an n such that f^i(n) = k for some i ≥ 0: Let n0 = k. If n0 is even, set n1 = n0/2; otherwise, set n1 = 3n0+1. Let n2 = f(n1) (apply f to n1). Continuing in this way, we construct a sequence ni decreasing to 1. Let n be the first ni that satisfies f(ni) = ni. Then f^i(n) = k where i is the number of steps taken. To see that n exists, we note that each ni satisfies ni ≥ 1, and so the sequence ni must eventually reach 1. Moreover, since f^i(n) is strictly decreasing, there can be at most one ni satisfying f(ni) = ni. Therefore, for all k ≥ 1 we can construct an n such that f^i(n) = k for some i ≥ 0, which shows that f is surjective. Since we have shown that f is injective and surjective, we have established that f is a bijection from the positive integers N to the eventual outputs {1,1,1,...} of the Collatz function. Therefore, the sets N and {1,1,1,...} have equal cardinality, which proves the Collatz Conjecture. QED

[@edwardoliveira1748]

Dude has such a relaxing way of speaking!

[@mistycremo9301]

I feel that this problem may be easy enough to find a counter example for if you could search for loops without testing any number. Maybe try plugging in 3(2x)+1=x, or other equations that show the process that might occur to any number, and if you find that a number that would, following the rules of the conjecture, normally be able to follow such a pattern, you would find a loop that hopefully does not end in 1.

[@JHashcroft]

Hey it’s been 5 years, did you manage to solve the Collatz conjecture?

[@subscheme]

lol@@JHashcroft

[@KirosanaPerkele]

Surely 6 years was sufficient to find your counter example. After all, you said it's easy enough. Let's hear it.

[@joaovitormatos8147]

This conjecture was in my maths test, it asked how many non-repeating primes was in the Collatz Conjecture. People failed because they thought 1 was prime

[@maldova]

Isn't this another unintentional asmr video? Professor's voice is so calming

[@tommykarrick9130]

I’ve watched this video so many times and this is so frustrating because it is so intuitive and clear that every number will obviously get to one and yet we can not prove it

[@tommykarrick9130]

awebmate the interesting thing about the problem is that it will always get to one and we don’t know why If it had a clear simple proof it probably wouldn’t be interesting

[@lukeblankenberg7371]

It seems to me (although this has probably been pointed out many times) that just because (3n+1)/2 > n we do not have to show that there are more n/2's in order to reach 1. What we have to do is show that it hits some power of 2 (as then it will immediately go to 1)(the last 1/2 should be fine as hitting a power of 2 or the next power of 2 is equivalent for the purposes of going to 1). The issue then is whether this is the limiting case (all cases with more n/2's in them before a power of 2, so that there aren't equal amounts of them, will all go to a power of 2 if this one does/they will eventually goes on a track that goes to this alternating between even and odd sequence that we might be able to show goes to a power of 2, or if some do not necessarily do so and so go to infinity). However, I have a nasty feeling that actually showing this is pretty hard. But I thought I might as well throw this out there.

[@jeremybuchanan4759]

"If they never get to earth, of course, they're not hailstones."

[@michaeldeierhoi4096]

😄😁😆

[@medexamtoolsdotcom]

I found a collatz calculator online and right off the bat found 2 numbers that last a long time. 447 for 80-something and 447123 which lasts for 138 iterations.

[@thesheq5023]

If you look at (3k+1)/2. It is consistent that after some trials you will find a number that fulfills the request of (2^n). So really you’re looking for a number that doesn’t satisfy the equation (3k+1)/2 = 2^n + 2^q + 2^..... so find a number that when put into (3k+1) isn’t a series of 2, however i don’t think one exists...

[@drenz1523]

Here is something I figured out: If the numbers will eventually end in a loop other than the 1-4-2 loop, it cant be in the pattern odd-even-odd-even... and so on, ending with an even and go back to the same odd number at the start. This might help!

[@dmtc6913]

I did some extremely precise calculations and all and here are the results. There's an infinite quantity of powers of 2 and of numbers that lead to a power of 2. Every single step you take with any number is literally just playing with fire and in the end you will get burnt.

[@drenz1523]

@@dmtc6913 so you solved it or somethin?

[@dmtc6913]

@@drenz1523 My previous post included the full proof. There's no way to avoid the infinite number of death traps forever. Unless you could end up in a loop other than 1 4 2 1. Surely there's no such thing. My maths is too precise for that.

[@drenz1523]

@@dmtc6913 wym by death traps? all will eventually lead to 1? just because there are an infinite amount of numbers that go to a power of two, doesn't mean other numbers will eventually branch into them. it's kind of like the that one orchard problem that numberphile covered (Tree Gaps and Orchard Problems) : in an infinite lattice grid, there is a line goes infinitely far but will miss all the lattice points if its gradient is irrational. maybe a special number is that line, and all the points are powers of two.

[@dmtc6913]

@@drenz1523 I must admit that my comments were mostly tongue in cheek, sorry. However, I can assure you with the utmost confidence that all numbers are doomed from the start. I didn't have to do any maths, I just come equipped with that knowledge. And give it out for free.

[@lopata_of_death6894]

"I've become more powerful than any jedi."

[@moonman____]

Twice the pride double the fall

[@alexritchie4586]

It's amazing how this problem seems so intuitive. That there's just something about the rules that seems so obvious the numbers will tend back to 1, but it's like having the solution on the tip of your tongue. Intuitively it's true, but you can't quite see why, despite the simplicity. I can see why people become obsessed with it.

[@svanse45.56]

This fascinates me soo much you wouldnt believe

[@lilcriz9187]

And this is, why I love mathematics :)

[@Verrisin]

I see why it tends towards 1: 2>1.5+c (for all big numbers and all small are checked) but I wonder what keeps it from ever entering a loop...

[@kinyutaka]

The numbers in the paths are on separate moduli. Take the number 7, and think of it as 128x+7 or 7 mod 128. Multiply by 3, add 1 and divide by 2 to get 192x+11 or 11 mod 192. Do that again, and you get 288x+17, then 216x+13, and 81x+5. Furthermore, you can cut out parts of these mods to show that the paths are part of even smaller moduli: 11 mod 192 is part of 11 mod 64. 17 mod 288 is part of 17 mod 32. And 13 mod 216 is part of 5 mod 8, leaving 5 mod 81 to be 2 mod 3 Because of this, we can clearly show the direction of the path along the branching tree of Collatz, and show that the number 7 (which we already knew) goes all the way to one. Generalizing it, we can show that any number above the point where we can verify has a matching pattern, a matching path, that lies within the numbers we can verify, and thus must also drop below itself to a number that has been verified, thus proving that all numbers, without exception, drop down to 1.

[@javierantoniosilva8477]

6:30 "People know all these things". Thats the awe I feel about maths in general, particulary in my undergraduate courses, like measure theory or abstract álgebra.

[@bengalbasi4753]

The " collatz behavior " is first shown when converting an odd p into even using p+1 and repeating collatz process infinitely. This collatz behavior originates from 1(p)+1. This exact behavior is observed using 3p+1. But loses using 5p+1.

[@TheReligiousAtheists]

1:26 I always knew no one uses multiplication tables!!😃

[@wotsac]

No, he just broke it down into something that fit in the (US) multiplication tables.

[@chizzicle]

I feel like there's been a video about this here before, but considering how many maths videos I watch, it's entirely possible that was someone else's video, especially since some of the people I watch occasionally appear here as well

[@DrGerbils]

Yeah. I remember seeing this recently as well, but a search for "collatz" doesn't turn up any videos I've watched recently.

[@NoriMori1992]

Pretty confident they've never ever covered this.

[@tamiratgebremariam8983]

If one can arrive at a power of 2, or power of power of 2, etc, then one is sure to arrive at 1 eventually. Categorization of the evens based on their 'proximity' to the powers of 2 could help.

[@dialecticalmonist3405]

Just when I thought I found my favorite math thing, another math thing comes along. This is my favorite math thing.

[@DiegoTuzzolo]

dis anyone notice the fibonacci sequence on the 7's tree? (quantities of numbers going down) 1 1 2 3 5 then the sequence ends

[@AaronHollander314]

Rule #1 has a lowering effect on any number and outputs an even number a certain percentage of the time. While rule #2 increases the number, it always produces an even number which is reduced further. Seems obvious that the number would tend downward. One step forward and two steps back. The misdirection is that you're tripling the number half the time, that's simply not the case.

[@rajeevbagra5276]

Agree completely.

[@legendgames128]

But if there was a number where the steps you take are like this: (3n+1)/2, (3n'+1)/2, (3n''+1)/2,... keep going like that forever, then the number would never reach 1.

[@daringcuteseal]

His voice is so soothing...

[@MrZooganopolos]

Very cool!

[@michaelbauers8800]

Is it just me, or is the quality of comments on this one weaker than usual? ;)

[@oluwatoyinokwunwa4989]

Michael Bauers Noticed that too. Thought I would see more educated comments.

[@medexamtoolsdotcom]

Roses are red, Violets are blue, 3n+1, Is eventually 2. Or maybe "Or divide it by 2".

[@jameshuddle4712]

(raises champagne glass) I found 21 to be amusing... But 27 simply did not know when to stop!

[@ra99nano21]

1:56 lol! It's weird to understand his surprise

[@Viewer2812]

"I can keep doing this and build up my tree" I love this line so much.

[@michaelepugliese7833]

It is an absolute fact that all positive integers resolve to a multiple of 5, or a power of 2. I can easily show this beyond any doubt. As often as not it can take a many iterations to reach a multiple of 10 that is divisible by 2 many times, sometimes immediately iterating to a (2^n)*5 number, and sometimes reaching another cycle, through various ending digits, to get another multiple of 5 ending in a zero, iterating down again by dividing by 2. The beauty of this is that you cannot avoid ending up with a multiple of 10 which is divisible by 2 all the way to a number that is represented by (2^n)*5, which will always iterate to a power of 2. At any given moment a power of 2 itself can magically appear as well. It seems airtight to me, and I would welcome someone finding a flaw in my method. I do not have the advanced mathematical skills to submit a formal proof of my findings, but the patterns are so obvious that someone with those skills should be able to do some amazing things with my findings. I would love to collaborate with such an individual.

[@griffontheorist6975]

I do recognize 5 as an important root number in 3x+1, but I don't know if understanding this is enough to prove the Collatz Conjecture to be true. The first even number to take an even number steps to be odd will always be a power of 2 (4 takes 2 steps, 16 takes 4 steps, 64 takes 6 steps, etc.) and the first even number to take an odd number of steps to be odd will always be a power of 2 * 5 (10 takes 1 step, 40 takes 3 steps, 160 takes 5 steps, etc.). However, this is also true for the modified rule 3x+7. Even though 5 and 1 are still essential roots, this all gets thrown in the 5-11 loop. This logic by itself does not explain why the 28-14-7 loop exists. However, I think you make an interesting point. 3x+7 seems unusually stable. The Collatz clone loop is required and unavoidable, so it's kind of cheating to point to that as the "exception". I don't believe there are any other loops aside from the 5-11 loop. Apparently this loop must be huge because it swallows a bunch of negative numbers too. However, looking over 3x+13, my initial ideas on roots fall apart. Starting with 16 is a little weird, then 3 *3+13 gives you 22, so the first root is 11? But 5*3+13 gives you 28, the first divide by 2 number, so 7 is a root? 3x+13 is also insane, there are a lot of loops and whatever the heck is going on with its septuplet loops.

[@fyfferguy]

To the OP: What is the source of the graph I see in the "collage" at 2:45? It's the one in the bottom center of the screen with the red dots on white background. This is exactly the data set that I have been playing with since I first heard of this problem in my college days over 25 years ago. I'd never seen the data presented that way before, and then I saw "my" graph in your video! Thanks in advance! (and hopefully someone will see this ... :( )

[@russellgokemeijer4269]

I have seen that graph on the Wikipedia page for collate conjecture and I am sure they list a source

[@WarpRulez]

Couldn't you build the tree in the opposite direction, ie. from the bottom up? In other words, start from 1 and do the reverse operations, always branching out when you can do both operations to the number.

[@pfeifenheini]

Yes you can. But it doesn't help.

[@WarpRulez]

Treufuß It helps building a complete tree up to a certain depth, which will then help you with further calculations.

[@WarpRulez]

***** Uh, no. You do the reverse operations, branching out every time you can do them both. In other words, you build the actual tree, but from bottom up.

[@ITR]

I think I saw a program that did that when I searched for more videos on the subject on youtube earlier

[@MaxRay16]

+WarpRulez You would only be getting even numbers

[@qwertteppig2272]

Beautiful conjecture

[@ImNotYo.u]

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.

[@adityakhanna113]

2:50 to the right, it's not a paper, It's an XKCD comic.

[@YoHoOMirster]

it's comic on the collatz conjecture

[@NoriMori1992]

Well-spotted! Don't think I'm familiar with that page, I'll have to Google it later!

[@rohanpandey2037]

lol I saw that too

[@GlobalWarmingSkeptic]

Extremely interesting! Just going over it I can already kinda see why this is a problem. You would have to find a set of numbers either where 3n+1/2 appears more than n/2, or you'd have to find a set of numbers that creates its own tree, which would probably be an extraordinary large number considering that, as numbers increase, it's less likely that you'll get a cyclical sequence.

[@alantyte3317]

A cyclical sequence is inevitable as there are only six rules for finding the pivot of the 1-chain which effectively boil down to 3. The length of the cycle triples at every step up. The numbers do get huge but are consistent. Easier to talk directly.

[@squashedoranges7949]

I would love to visualize this!

[@lucasandrescosta391]

Really fun problem to code as beginner programmer and try some numbers out.

[@ClaudioBrogliato]

The only pattern I see is that powers of two with even exponential, once you subtract one, become odd numbers that can be divided by 3.

[@cuentafake140]

If you have odd powers of two and add one, that number is divisible by 3

[@giridharlprabhu8939]

This is a result in number theory which is more formally defined as 4^x-1|3

[@official-obama]

@@cuentafake140 subtract* Also the only odd power of two is 1

[@cuentafake140]

@Orion Hunter Oh, I meant to say that 1 + 2^(odd number) is always divisible by 3, for instance: - 2^5 + 1 = 33 - 2^7 + 1 = 129 - 2^15 + 1 = 32769 My english is pretty mediocre at best so... sorry z.z

[@52.yusrilihsanadinatanegar79]

2:36 I've done checking 10 ^ 100 + 1 on my pc and it still got 1. A googol plus one. Down to ONE.

[@inx1819]

how many steps?

[@Quantris]

@@inx1819 It only takes 2173 steps. What's kinda interesting is that 10^100 + 2 and 10^100 + 3 also take this many steps

[@inx1819]

@@Quantris yeah I made a program for that as well, it was pretty fun. got the same results

[@austinaxley81]

I could be thinking about this incorrectly but, it seems to me once you touch a power of 2 you're guaranteed to fall down to 1. So the question is ultimately, is there a positive integer you can insert into this function that, no matter the iteration, would never result in a power of 2. Simple logic, it seems to me, indicates that since every positive integer can be expressed as some combination of powers of two, that this would be impossible. I could be down a rabbit hole though, I'm not a mathematician.

[@Knurf]

The next number with more steps than 63,728,127 has 1 step more and is double of it. And even though that 63 million number has 949 steps, the lowest number with over or equal to 1000 steps is 1,412,987,847 with exactly 1000 steps and is way bigger than that. And the number with the most steps that is under 4 billion is 2,610,744,987 with 1050 steps.

[@wakeupnawaf]

why do they always draw on brown toilet papers?

[@leepoling4897]

That's clearly not toilet paper

[@EGarrett01]

Who poops on brown paper?

[@DjSamvy]

Remind me to never use a bathroom where you live if that's what the toilet paper is like

[@marianpalko2531]

iiDioxide Brown toilet paper, to use your vocabulary, is the iconic symbol of Numberphile.

[@gustavrsh]

it's cheap.

[@RedDevilWOT]

Hi.I have a question whether at always 3n + 1 / 3n-1 we will come to a cycle from 4-1 / 2-1 or in the second case there are exceptions. I found the number 123 in the first case, it comes to one, and at 3n-1 it loops on the number 17. Is it possible to get an answer?

[@rodrigoaceves9020]

this is a really cool way of finding prime numbers

[@chonchjohnch]

I decided to try a graph approach to this, specifically digraphs, every vertex has two incoming edges, one representing 3n+1 for some n and one representing n/2 for some n. Additionally each vertex has one outgoing edge which goes to a vertex representing the collatz function being applied. Basically every number has two incoming edges and one outgoing edge