I tried to use this conjecture for a compression algorithm once. My thinking was, if you take a very large number, and keep dividing it, you'll eventually get down to a smaller number, which takes up fewer bytes. If it's even, divide by 2. If it's odd, subtract 1 and divide by 2. Do the opposite when expanding it again. Do this recursively, you'll eventually convert a multi-GB file down to a mere hash string of bytes. It works no problem, in theory. The problem is, of course, expanding it again. The information required to know when the original numbers were odd or even is basically the same amount of data the original file was. So basically, all it was was a weird encryption algorithm.
I thought I had proved the Collatz Conjecture the other night. I couldn't believe it! Little old me had succeeded where countless brilliant mathematicians had failed! I must be a genius of the highest rank, I thought! Then I noticed that the very first line of my proof was totally wrong. Oh well. At least there was that brief moment of glory in my addled brain. I take what I can get.
I too, like probably many numberphiles, often dream of stumbling my way to solving an a open problem in mathematics. It would be incredible! I think it would most definitely require divine intervention or inspiration to solve something that has escaped the reach of professional mathematicians who dedicate their life, to take that immense creative leap that finally let's you see the problem in a way that is solvable. Has this happened before? Has an amateur solved an open problem the escaped the professionals? The best candidate I can think of is Ramanujan, who was self taught and isolated from the math would, iirc, and sent G.H. Hardy a number of incredible results?
You could colour it in 4 colours, but it might be impossible to colour one string in one colour, because sometimes a string overlaps another string and one string has to be drawn in 2 parts.
That's what jogiff meant by exclaves. An exclave is basically two regions which are disconnected but must have the same colour. This additional restriction means that a 4-colouring might be impossible.
What makes this problem so interesting is that given how the function jumps up and down one would think there should be infinite loops, even if they are individually very rare. If there are no other loops then there must be a reason and given the simplicity of the function it should be easily discernable. Must drive number theorists nuts. 8O
To add to that, I think, that any loop would have to be outside the main tree (the tree with root 1). Reason is that applying the rules forwards (i.e. if n is even, next is n/2, if n is odd...) never diverges to two branches. To have a loop or cycle in the graph presented here you would need a node diverge in the forward direction, but they can only diverge in the backward direction. 1-2-4 is the only possible loop within the presented graph. Correct me if I'm wrong. That being said there could be numbers a, b, c, ... that form a closed loop but are separate from the main tree. It would be that not all n reach 1, but also don't go off to infinity. Could there be branches coming out of the loop?
@@josevillegas5243 There are 2 ways the conjecture could be false, either there exists a loop other than 1, 4, 2 or there is a sequence that eventually goes to infinity. In either case there then exist a load of other numbers that drop down into those sequences. At a minimum there are all numbers that are a power of 2 times members of these sequences. Since it is deterministic once you get in a loop you stay in that loop.
I came up with a triple formula that makes a similar, but different pattern. If divisible by 3, then n/3. If n-1 is divisible by 3, then 4n+2. If n-2 is divisible by 3, then 2n+2. You end up with two completely separate branches that converge into 2 and 10, but contain every number, just like Collatz.
I will be the first to post this today. I've found a marvellous proof to the Collatz Conjecture, however there is not enough space in the youtube comment section to submit it here.
This reminds me of cellular automata and other systems that develop into something very complex and seemingly random from simple rules Stephen Wolfram talked about in his "A New Kind of Science".
Yes! The issue is that we don't really know which kind of complexity class the collatz conjecture is in.... - On the inner digits is repeated multiplication by 3 in base 6, which scrambles everything so it's a chaos-dominated random cellular automata.... - On the outer digits, it's the loss of the last digit every time the last number is even, eating up digits, so it's a decay-to-fixed-state cellular automata.... - There's no proof that it's impossible to produce complex, localized structures like the famously Turing-complete rule 110 either....
Interesting conjecture. If you want to know the number of steps using Excel, enter a number in cell A1, then enter the following formula in cell A2: =IF(A1="END","END",IF(A1=1,"END",IF(ISEVEN(A1),A1/2,(A1*3)+1))) Pull down the formula ~2,000 rows and enter the following formula in cell B1 to count the steps needed to get to 1: =IF(B2=">15 DIGITS - CALCULATION INCORRECT","NULL",COUNT(A:A)-1) Bear in mind that Excel does not calculate correctly if any number is >15 digits, so add this formula to cell B2 that will tell you either the largest number in the data set, or if you are over: =IF(MAX(A:A)>999999999999999,">15 DIGITS - CALCULATION INCORRECT",MAX(A:A)) I believe the maximum number of steps that can be found using Excel is 1,228 for 75,128,138,247. Cheers!
Beatiful! I heard about the Collatz Conjecture before, but this "sea weed" image is truly mind-boggling! It shows how nature and math are connected to each other.
Collatz Conjectures, Mandelbrot Fractals, Bonini's Paradox, Solar Accretions, so much of our complex universe comes from simple origins, who wants to bet that life and consciousness are just as simple?
You seem to have it backwards. There is nothing simple or simplistic about these maths. They are what you might call infinite amounts of information condensed into a single phrase or word of semiotic linguistics. You only think that they are simple because you view them as their starting point. An acorn is very simple, but it contains all the vast amounts of information to form a tree, given the proper context. When you view the acorn and the tree as the same thing, expanded across time, you start to realize that it's not simple at all, but rather it is the single most complex system in all of existence. Other things in life are the same way. Human traditions, for example, seem simple and silly at times. But when you realize that Tradition is actually Democracy extended through time, handed down from generation to generation, you start realizing that traditions are vastly complex things which involved thousands or maybe even millions of people. A pencil is another example of something that is remarkably complex, but is a very simple object. It would be very difficult for any human being to produce a pencil by himself, without anybody helping him. It takes woodcutters and coal miners and all sorts of manufacturing plants just to get the raw materials and turn them into a pencil. Be very cautious not to mistake simple DESCRIPTIONS of something for the thing itself being simple. Being able to encode information about an object in very compact coded ways doesn't mean the object itself is simple.
Triumvirate888 - indeed. Feynmann said if you knew the names of all the birds you still wouldn't actually *know* anything about those birds. You just simply memorized the names. You have no wisdom. Identifiers, names, these things are not real wisdom, just rote knowledge.
This video is a piece of art. Also with this illustration you can explain this easily: The Cinjecture is if all of these branches end up in the point of the bottom or if there are some loops.
For a given odd number n_i you have the sequence of even numbers n_i^k for all k in [1,2,3,4,...]. Thus, 3n_i+1 = 2^l*n_j (because you are dividing by 2 each time you have an even number). That is, n_i will join onto the n_j branch of the collatz tree (the tree of all numbers spanning from 1 following the collatz conjecture). The collatz conjecture then holds if all odd numbers map to an odd number in the collatz tree. In the example in the video, you have the sequence [13, 5, 1] because 3*13+1 = 2^4*5 and 3*5+1 = 2^5*1. Therefore, what is more interesting is not the links between the numbers, but the connections between the odd number at the root of each even number sequence.
2^n series as main trunk apply (n-1)/3 rule on it will get prime or composite number by apply n/2 rule between too, time each number in 2^n series again repeat same (n-1)/3 rule will branch out to every integer number, by trace back every n reach 1 by reverse (n-1)/3 rule to 3n+1 rule.
I've been thinking about this for a while and I think the Collatz Conjecture is totally true. Every finite number does go on down to 1 at some point following the Collatz rule. Here's why this happens. Look at it in binary. 13 = 1101 in binary. Because the last digit is a 1 it's multiplied by three and has one more added. This ALWAYS makes the last digit 0. It also pushes all of the other 1's farther up, and usually eliminates one of them. Sometimes it eliminates two or three. 40 = 101000 There's the proof in the pudding. Now, the computer just starts knocking off 0's until it finds another 1. 20 = 10100 10 = 1010 5 = 101 Here we do the 3x+1 again. 16 = 10000 And once it knocks off all the zeroes this time we get to 1. I'll have to do some more looking into it, but I think I may be onto something.
I feel like it would be difficult to prove that all numbers will do this. I feel like somewhere there could be an odd number which, when doing this, it ends up with 1 0 to cancel, then the new number with only 1 0 to cancel, and so on, allowing it to be infinitely big. I feel like proving this would be equally difficult as proving the conjecture (especially considering that I kinda doubt no-one has thought of this until now)
19 = 10011 (3 ones) 19 * 3 + 1 = 58 = 111010 (4 ones) When 19 is multiplied by 3 and you add 1, you actually increase the number of ones. Additionally, 58 ends in only one zero. When you eliminate the zero, you get 29, which is greater than 19. There might be a number that does this forever, increasing to infinity. There could also be a cycle, where eventually you end up with the same number that you started with.
If you assume a number is the smallest number that it's sequence doesn't terminate, you can tell it must be a 12*n+3 or 12*n+11 number because if it were anything else it would either divide 2, which means there's a smaller number with a sequence that doesn't terminate, or multiplying it by 3 and adding 1 would result in a number that divides 4, which would also mean after dividing by 2 twice you have a smaller number than you started with with a sequence that doesn't terminate...
No one has ever seen Andy Serkis. Even the interviews he does are CGI. For all we know, he's a hyper-intelligent goldfish in a bowl, connected to a brainwave-interpreting computer through electrodes.
I find it interesting, cause if you look at the picture, seeing how every number turns from the previous one, the entire sequence is trending toward more even numbers. Yeah, odds are sprinkled in everywhere, but the whole thing is turning clockwise. Maybe thats because of the limit of numbers used tho...
Collatz conjecture goes to 1 because there is more numbers divisible by 2 then there is odd numbers. So series like /2/2 happen more frequently, and a chance to hit number divisible by 12,16, etc increases over time (because every next number is different from previous, you can transfer them to base 2 or 3 to see why)
I loved the coloring in this video! The appreciation the mathematician had for this visual rendering of this problem was so cool. Total +1 for having a menora Brady, what an unexpected treat : )
It would be interesting to compare this tree visualisation to trees of other similar "systems". Then you could see what you can really conject from having a "weird" tree vs. an "orderly" tree. Maybe some trees from systems where not everything is connected to one are weird too or some systems connect everything too one with an orderly tree.
Is anyone else really bothered by the structure of those tree examples? Like at 3:43 the number 3 is connected to 5 when it should be connected to 10. 5 is connected directly to 8 when it should have to go through 16 first. That isn't on purpose right? Maybe it was just explained confusingly??
Mage of Void is right imo. Connecting 5 to 8 makes as much sense as connecting 8 to 2: it's skipping a step. Sure, 5 will go to 8 as 8 will go to 2 but 5 gets to 8 through 16. Same with 13 to 20. It goes through 40 so should connect to it. The diagram is consistent in combining an odd step with its subsequent even step so no technically wrong as a schematic but I don't see how that is helpful.
N.J.Wildberger uses the Collatz conjecture to puncture ones naive notions of just how good infinite processes and definitions of convergence are. He says in analysis you only deal with easy sequences of numbers.
If n is odd, 3n+1 is even, so one will immediately divided by 2, so one could make the step (3n+1)/1 before checking polarity again. The tree would then be straighter with know evens removed.
When you have an odd number and multiply it by three, you will have an odd number. Add one and you get an even number. So then you divide it by two. For convenience, some people make (3n+1)/2 into one step. 5 goes directly to 8 instead of 16.
7 лет назад
Beautiful video, thanks. However, 3:07 the trees are a bit wrong if you follow the collatz conjecture rule, right? 3 should connect to 10 (and not to 5) 13 to 40 (and not to 20) 21 to 64 (not to 32) 85 to 256 (not 128) Hmmm, looks like it is mentioned in 4:24 (for odd number do 3n+1 and then divide by 2). I am wondering, why didn't the artist use the original collatz conjecture rules but slightly altered them this way? To make the tree look more "compact" with branches closer together, or? Thanks.
All you have to do is prove that with any odd number, when following the steps, you eventually get to a power of 2. Idk if that's hard to do but it's a start if they don't have one.
I have some information to spread (ifn't it already exists) So first reverse the Conjecture so /2 is *2 and *3+1 is -1/3. Then with those equations you can apply either *2 or -1/3 to even numbers, and you can only *2 odd numbers. now if you do a tree with only 2^x you quickly realise that only x in this case must be even for it to be a choice between *2 or -1/3. And another cool thing is that if you add your current even 2^x with it's -1/3 result you'll get the next in the lines result. Illustration: 1 4+1=5 16+5=21 /\ /\ /\ 4 -> 8 -> 16 -> 32 -> 64 . . . The equation for that being 2^x + (2^x-1)/3 = (2^(x+2)-1)/3. I hope I provided something interesting.
If a n = 2^m, then applying the algorithm to n WILL give a sequence that achieves 1 at the iteration m. If n is even & NOT an integer power of 2, then there exists some number of iterations p such that after p iterations, the resulting number is an odd number. Therefore, the question we are interested in is equivalent to the question, "if n = 2m + 1, then can we always reach some power of 2 by having 3n + 1?" 3(2m + 1) + 1 = 6m + 4, and this is a power of 2, then 3m + 2 is a power of 2. We want 3p + 2 = 2^q with natural solutions p & q. If a starting number can achieve a solution p to this equation, then that number achieves 1. This is truly a question about modular arithmetic.
That is amazing! I did my thesis making a model of seaweed (Ulva lactuca) growth. I love how something organic-looking can be "created by" such a short mathematical formula... And I'd totally love to have this as a poster on my wall. Guess I'll have to get the coloring book and start getting creative ^^'
Not "created by", just "described by". Mathematics is mankind's way of giving linguistic adjectives to nature, describing things and events in the abstract and extrapolating outwards to gain ever-more precise logical understanding of those things and events. I'm sure you already knew that though, since you did a thesis on it. I just wanted to make it clear for others who might read this and mistakenly think that mathematics creates things. If people like Stephen Hawking can make such a silly blunder, then anybody can.
Things like this, and most certainly the prime numbers, are concepts people will never understand; the intricacies of the positive integers transcend human thought. As Leopold Kronecker was quoted as saying, ""God made the integers, all else is the work of man."
mvmlego1212 I have not checked it and i have no idea how they did it but i would guess by the value of each branch where they intersect. could just be random though.
It's something wrong with this image. The Collatz three branches off at all numbers of the form 6n+4 and only those. I call them nodes. A node creates a vertical branch (2n-branch) and a horisontal branch ( (n-1)/3-branch). This means that the nodes are 10, 16, 22, 28 and so on. It is simple to show that all numbers will end up in a node. It is also simple to show that all the odd numbers are located as the first number on a horisontal branch (and hence there is a one-to-one relationship between all odd numbers and all the nodes, and hence all the horisontal branches) (If you can show that all the nodes are connected you have hit the jack-pot...) There are three types of nodes, the ones of type 18k+4 (this is type 1), 18k+10 (this is type 2) and 18k+16 (node type 3). Nodes of type 2 creates a branch that does not have any new nodes. Type 2 nodes are 10, 28, 46 and so on. Only type 1 and 3 creates horisontal branches with new nodes. And there is a pattern to what type of node the next node is - it is not "chaotic" - even if it looks chaotic... The long branch at the bottom of the figure does not have any new branches branching off - this means that this is a type 2 horisontal branch. This branch must start with an odd number, all the rest are even. One such horisontal branch of type 2 is 3,6,12,24,48, ..., 3*2^p - this branch originates from 10 (the node), but there are no new nodes on this brach - no number in this branch will ever equal 6n+4 and all of them except the first number are even (as are the case with all horisontal branches). This means that this branch in the figure shall always turn clockwise. But it clearly doesn't - it turns a little bit in this and a little bit in that direction..???... Actually it turns mostly anti-clockwise, which means that there are mostly odd numbers in the branch, which is impossible, since branches which do not have any new branches only have even numers in them - the numbers are e*2^p, where e is an even number in the series 3, 9, 15, 21, ... (the starting number of the branch). So 3, 6, 12, 24 is one such branch and 9, 18, 36, 72 is the next and 15, 30, 60, 120 the third type 2 branch and so on. Hence the computer code that makes this figure has some type of "randomness" written into the code - making it look more chaotic than it really is. I hope that they can remove this randomness and show a "pure" one - I like playing with the collatz conjecture and would love to see it. By the way, type 1 and type 3 nodes creates new nodes, and there is a simple pattern to what the new node is and what type of node the next one is. But showing that all the nodes are "connected" is the difficulty.....
odd has to be preceded by even, and followed by even. So for each 3n+1 there is a divide by 4, 2 before 2 after. So 18 > 9 > 28 > 14 It has to reduce to one. what am I missing? if you repeatedly multiply by 3 but divide by 4 I'm guessing it'll get smaller. and smallest positive integer is 1, or maybe zero?
Why does it have small figures like sucker of an octopus under lines? It leads thinking of underwater creatures. And why do branches from even numbers overlap the lines from odd numbers?
I have a solution. Since every odd will have 1 added to it every odd will become even next turn. The only way to get down to 1 is from a number with a root of 2 like 2^3 or 2^n. So therefore, every odd number will become even and then odd again and over and over until the even is a root of 2. For example, 13 goes to 40 which goes to 20, then 10, 5,16, 8, 4, 2, 1.
Seriously, though, not all even numbers are a power of 2. How would you prove that it goes to a power of 2 at one point? How do you know if there isn't a number that doesn't go to a power of 2 at all?
those trees remaind me of philogenetic trees that show the common ancestor of several species. Wonder if we could ever apply this conjecture to predict patterns of evolutionary process, or something like that. sorry for bad english
What would happen if you add # of iterations as a third component lets say going up and duration as in how long since each iteration was calculated as thickness of the branches? Would it look like a tree with the absence of wind and a perpendicular constant sun?
So what he's saying is that there's no visible pattern? You could say it seems arbitrary for all we know. But isn't the premise/goal in itself arbitrary and the result to be expected or at least not surprising?
I think that is similar to proving that you can't win in a casino if you play long enough. The probability that the next number in sequence is odd is lower than the probability that it's going to be even. I imagine that the sequence length is exponentially smaller than the starting number. But I'm probably wrong :)
"Every number less than 10,000" Is every branch extended until it reaches the first five-digit number? Or is every branch explored until every four-digit number has been found? I'm guessing he meant the former, because I once mapped out a tree for *every* two-digit number, and there were a couple of outlier branches significantly longer than the rest. I don't see any long outliers in that image, which I would expect would have many more dramatic examples if including all four-digit numbers.
i have discover something amazing that like in abstartc algebra we have feilds like natural number is an subset of whol number and whole number is a subst of integers etc then in numbers like you see that prime and composite number is a subset of even and odd numer
Because as soon as you hit an odd number you make it even, so in the best case scenario for odd numbers 50% of them would be odd and 50% even. This is obviously not the case as when you hit an even you can get either an even number or an odd one next.
Hey can you guys do a video on TREE(3)? I'd like to learn more about its origins, the TREE function itself, and how much larger it is than Graham's Number because I know its quite a significant gap. Thank you!!
I tried to use this conjecture for a compression algorithm once. My thinking was, if you take a very large number, and keep dividing it, you'll eventually get down to a smaller number, which takes up fewer bytes. If it's even, divide by 2. If it's odd, subtract 1 and divide by 2. Do the opposite when expanding it again. Do this recursively, you'll eventually convert a multi-GB file down to a mere hash string of bytes.
It works no problem, in theory. The problem is, of course, expanding it again. The information required to know when the original numbers were odd or even is basically the same amount of data the original file was. So basically, all it was was a weird encryption algorithm.
after watching numberphile for this long, it seems there are 2^60 "most famous" things in math.
I'd say more like 7 billion, because everyone has their own favorite thing that is the most famous to them :)
IceMetalPunk some people may share things though so it must be MUCH less than 7billion
not even close to 7 billion, you dont think at least 1 overlapping of favorite concepts has occured?
But maybe they are all collapsing to one...
60²
I thought I had proved the Collatz Conjecture the other night. I couldn't believe it! Little old me had succeeded where countless brilliant mathematicians had failed! I must be a genius of the highest rank, I thought! Then I noticed that the very first line of my proof was totally wrong. Oh well. At least there was that brief moment of glory in my addled brain. I take what I can get.
Can you share the logic?
Never have i felt so understood. I too, have tried my hand at it...
You gave it a go! You gave it a Parker! That's the important thing.
I too, like probably many numberphiles, often dream of stumbling my way to solving an a open problem in mathematics. It would be incredible! I think it would most definitely require divine intervention or inspiration to solve something that has escaped the reach of professional mathematicians who dedicate their life, to take that immense creative leap that finally let's you see the problem in a way that is solvable.
Has this happened before? Has an amateur solved an open problem the escaped the professionals? The best candidate I can think of is Ramanujan, who was self taught and isolated from the math would, iirc, and sent G.H. Hardy a number of incredible results?
@@kaziaburousan166 Kazi, I'm sorry, but I have forgotten any details of my false proof. I'm sure it was not salvageable.
Apart from all this, Lothar Collatz was one of the kindest persons I've ever met. Pure genius! R.I.P.
Has he put it on a t-shirt yet? I'd wear it.
fatsquirrel75 - yea, it's beautiful
Mathsgear pleasssse!
I'd like a print of that, put it on me wall!
Or a print. I'd have it on my wall in a flash.
i would love to jave a print of it indeed
Should have used only 4 colours while colouring it in.
I'm not sure the overlaps would allow you to use only 4 colors without coloring neighboring branches the same way.
He was probably refering to the four colour map theorem, check out that vid
The four color map theorem doesn't apply if there are exclaves so TiagoTiago is right. This map might very well not work for the four color theorem
You could colour it in 4 colours, but it might be impossible to colour one string in one colour, because sometimes a string overlaps another string and one string has to be drawn in 2 parts.
That's what jogiff meant by exclaves. An exclave is basically two regions which are disconnected but must have the same colour. This additional restriction means that a 4-colouring might be impossible.
I'd love to see this 3D-printed.
What makes this problem so interesting is that given how the function jumps up and down one would think there should be infinite loops, even if they are individually very rare. If there are no other loops then there must be a reason and given the simplicity of the function it should be easily discernable. Must drive number theorists nuts. 8O
To add to that, I think, that any loop would have to be outside the main tree (the tree with root 1).
Reason is that applying the rules forwards (i.e. if n is even, next is n/2, if n is odd...) never diverges to two branches.
To have a loop or cycle in the graph presented here you would need a node diverge in the forward direction, but they can only diverge in the backward direction.
1-2-4 is the only possible loop within the presented graph. Correct me if I'm wrong.
That being said there could be numbers a, b, c, ... that form a closed loop but are separate from the main tree. It would be that not all n reach 1, but also don't go off to infinity. Could there be branches coming out of the loop?
@@josevillegas5243 it's been proven that if a loop exists it'd have to be at least size 17087915 outside the main branch (of course)
@@josevillegas5243 There are 2 ways the conjecture could be false, either there exists a loop other than 1, 4, 2 or there is a sequence that eventually goes to infinity.
In either case there then exist a load of other numbers that drop down into those sequences. At a minimum there are all numbers that are a power of 2 times members of these sequences.
Since it is deterministic once you get in a loop you stay in that loop.
I came up with a triple formula that makes a similar, but different pattern. If divisible by 3, then n/3. If n-1 is divisible by 3, then 4n+2. If n-2 is divisible by 3, then 2n+2.
You end up with two completely separate branches that converge into 2 and 10, but contain every number, just like Collatz.
I am forever haunted by the way he says "uh-oh" at 1:18
I will be the first to post this today.
I've found a marvellous proof to the Collatz Conjecture, however there is not enough space in the youtube comment section to submit it here.
Mister Bateman are you by any chance a descendant of Fermat?
Mister Bateman I have an original joke that the margin is too narrow to contain
You are closer than you realise.
The proof is related to fermats theorum. IMO.
Shut up Fermat! You have caused so many mathematical headaches already LOL
*Bateman's Last Theorem* :)
This reminds me of cellular automata and other systems that develop into something very complex and seemingly random from simple rules Stephen Wolfram talked about in his "A New Kind of Science".
Thanks for the reference! it sounds interesting :)
Yes! The issue is that we don't really know which kind of complexity class the collatz conjecture is in....
- On the inner digits is repeated multiplication by 3 in base 6, which scrambles everything so it's a chaos-dominated random cellular automata....
- On the outer digits, it's the loss of the last digit every time the last number is even, eating up digits, so it's a decay-to-fixed-state cellular automata....
- There's no proof that it's impossible to produce complex, localized structures like the famously Turing-complete rule 110 either....
It measures a number's resistance to becoming a power of 2 based on 3n + 1.
That sounds like it is measuring "failure to exponentiate".
Basically a "general resistance" in a general environment.
WHY would you leave all the marker caps OFF while ONLY USING ONE AT A TIME 👊🏻😤
Yes, but why would you need to do that, if you're not constantly switching between them?
Monica, is that you ? :-)
I was thinking about this, too since those markers are fricking expensive.
... because he has gotten rich from youtube videos.
to divide the time taken by 3
The colors you chose for that seaweed illustration makes it look like something from a horror anime, like Tetsuo's power meltdown or something.
Interesting conjecture. If you want to know the number of steps using Excel, enter a number in cell A1, then enter the following formula in cell A2:
=IF(A1="END","END",IF(A1=1,"END",IF(ISEVEN(A1),A1/2,(A1*3)+1)))
Pull down the formula ~2,000 rows and enter the following formula in cell B1 to count the steps needed to get to 1:
=IF(B2=">15 DIGITS - CALCULATION INCORRECT","NULL",COUNT(A:A)-1)
Bear in mind that Excel does not calculate correctly if any number is >15 digits, so add this formula to cell B2 that will tell you either the largest number in the data set, or if you are over:
=IF(MAX(A:A)>999999999999999,">15 DIGITS - CALCULATION INCORRECT",MAX(A:A))
I believe the maximum number of steps that can be found using Excel is 1,228 for 75,128,138,247.
Cheers!
Beatiful! I heard about the Collatz Conjecture before, but this "sea weed" image is truly mind-boggling! It shows how nature and math are connected to each other.
Alex: "What do you think this is?"
Brady: "Collatz obvs innit"
Ooh! Something i've never heard about before. I'M A GONNA LEARN ME A THING!
Collatz Conjectures, Mandelbrot Fractals, Bonini's Paradox, Solar Accretions, so much of our complex universe comes from simple origins, who wants to bet that life and consciousness are just as simple?
They absolutely are. The universe is nothing but emergence, and it's beautiful in that :)
Layer upon layer of emergence creates some incredibly beautiful things indeed.
You seem to have it backwards. There is nothing simple or simplistic about these maths. They are what you might call infinite amounts of information condensed into a single phrase or word of semiotic linguistics. You only think that they are simple because you view them as their starting point. An acorn is very simple, but it contains all the vast amounts of information to form a tree, given the proper context. When you view the acorn and the tree as the same thing, expanded across time, you start to realize that it's not simple at all, but rather it is the single most complex system in all of existence.
Other things in life are the same way. Human traditions, for example, seem simple and silly at times. But when you realize that Tradition is actually Democracy extended through time, handed down from generation to generation, you start realizing that traditions are vastly complex things which involved thousands or maybe even millions of people.
A pencil is another example of something that is remarkably complex, but is a very simple object. It would be very difficult for any human being to produce a pencil by himself, without anybody helping him. It takes woodcutters and coal miners and all sorts of manufacturing plants just to get the raw materials and turn them into a pencil.
Be very cautious not to mistake simple DESCRIPTIONS of something for the thing itself being simple. Being able to encode information about an object in very compact coded ways doesn't mean the object itself is simple.
Triumvirate888 - indeed. Feynmann said if you knew the names of all the birds you still wouldn't actually *know* anything about those birds. You just simply memorized the names. You have no wisdom. Identifiers, names, these things are not real wisdom, just rote knowledge.
This video is a piece of art.
Also with this illustration you can explain this easily: The Cinjecture is if all of these branches end up in the point of the bottom or if there are some loops.
I never thought I'd want a coloring book this badly.
Can you color that tree with only 4 colors without a shade touching its self?
yes but it would not look 3d.
it's not a map, each line is continuous and goes under other lines. if it was shaded with only 4 colors it wouldn't look right.
For a given odd number n_i you have the sequence of even numbers n_i^k for all k in [1,2,3,4,...]. Thus, 3n_i+1 = 2^l*n_j (because you are dividing by 2 each time you have an even number). That is, n_i will join onto the n_j branch of the collatz tree (the tree of all numbers spanning from 1 following the collatz conjecture). The collatz conjecture then holds if all odd numbers map to an odd number in the collatz tree. In the example in the video, you have the sequence [13, 5, 1] because 3*13+1 = 2^4*5 and 3*5+1 = 2^5*1.
Therefore, what is more interesting is not the links between the numbers, but the connections between the odd number at the root of each even number sequence.
2^n series as main trunk apply (n-1)/3 rule on it will get prime or composite number by apply n/2 rule between too, time each number in 2^n series again repeat same (n-1)/3 rule will branch out to every integer number, by trace back every n reach 1 by reverse (n-1)/3 rule to 3n+1 rule.
I've been thinking about this for a while and I think the Collatz Conjecture is totally true. Every finite number does go on down to 1 at some point following the Collatz rule. Here's why this happens. Look at it in binary.
13 = 1101 in binary. Because the last digit is a 1 it's multiplied by three and has one more added. This ALWAYS makes the last digit 0. It also pushes all of the other 1's farther up, and usually eliminates one of them. Sometimes it eliminates two or three.
40 = 101000 There's the proof in the pudding. Now, the computer just starts knocking off 0's until it finds another 1.
20 = 10100
10 = 1010
5 = 101 Here we do the 3x+1 again.
16 = 10000 And once it knocks off all the zeroes this time we get to 1.
I'll have to do some more looking into it, but I think I may be onto something.
I feel like it would be difficult to prove that all numbers will do this. I feel like somewhere there could be an odd number which, when doing this, it ends up with 1 0 to cancel, then the new number with only 1 0 to cancel, and so on, allowing it to be infinitely big. I feel like proving this would be equally difficult as proving the conjecture (especially considering that I kinda doubt no-one has thought of this until now)
19 = 10011 (3 ones)
19 * 3 + 1 = 58 = 111010 (4 ones)
When 19 is multiplied by 3 and you add 1, you actually increase the number of ones.
Additionally, 58 ends in only one zero. When you eliminate the zero, you get 29, which is greater than 19. There might be a number that does this forever, increasing to infinity. There could also be a cycle, where eventually you end up with the same number that you started with.
ok that`s beautiful. i wanna make one now
Something so simple and dull can be made into something so complex and beautiful.
Beautiful! But now I lost even more hope that it gets solved in near future
If you assume a number is the smallest number that it's sequence doesn't terminate, you can tell it must be a 12*n+3 or 12*n+11 number because if it were anything else it would either divide 2, which means there's a smaller number with a sequence that doesn't terminate, or multiplying it by 3 and adding 1 would result in a number that divides 4, which would also mean after dividing by 2 twice you have a smaller number than you started with with a sequence that doesn't terminate...
looks like a sea coral! what a beautiful world.
My favorite conjecture 😊
Bravo! Visualizing the beauty and complexity of mathematical problems is vital for understandig them by our organic, biological mind.
That guy looks exactly like Andy Serkis
Tyler Matthew Harris I'd say he looks more like Michael Sheen.
his ugly , nerdy twin then.
+Antony Jones , damn! You're right.
No one has ever seen Andy Serkis. Even the interviews he does are CGI. For all we know, he's a hyper-intelligent goldfish in a bowl, connected to a brainwave-interpreting computer through electrodes.
we love the collatz conjecture!!!!
I find it interesting, cause if you look at the picture, seeing how every number turns from the previous one, the entire sequence is trending toward more even numbers. Yeah, odds are sprinkled in everywhere, but the whole thing is turning clockwise.
Maybe thats because of the limit of numbers used tho...
Collatz conjecture goes to 1 because there is more numbers divisible by 2 then there is odd numbers.
So series like /2/2 happen more frequently, and a chance to hit number divisible by 12,16, etc increases over time (because every next number is different from previous, you can transfer them to base 2 or 3 to see why)
The ratio between odd/even numbera in sequences is exactly 0.66(counted in program), so it appears twice as frequently as odd number do.
how far did you go ??
I loved the coloring in this video! The appreciation the mathematician had for this visual rendering of this problem was so cool. Total +1 for having a menora Brady, what an unexpected treat : )
at 3:06 why does 5 go into 8, 3 go into 5, 13 go into 20 and 21 go into 32
Please make a poster!
Brandon Sams, how about a program?
working on it, don't worry
Albert Zhang, I was here first!
Yay!! Tiff's in a Numberphile video
It would be interesting to compare this tree visualisation to trees of other similar "systems". Then you could see what you can really conject from having a "weird" tree vs. an "orderly" tree.
Maybe some trees from systems where not everything is connected to one are weird too or some systems connect everything too one with an orderly tree.
Is it not a bit obvious that 2^60 goes down to 1?
Joshua Green I think it was from a previous video where it said they'd computed that all numbers up to 2^60 go down to 1
It is proven, that every number up to 2^60 goes down to 1, including eg (2^60)-1.
Joshua Green I think they meant "All numbers up to 2^60 go down to 1"
A Kaizofan All numbers that can be written in the form 2^n, where n is a positive integer, will go down to 1.
(2^n)/2 = (2^n)/(2^1) = 2^(n-1)
It seems like these things will meander around but eventually they hit a factor of 2 and go to 1 pretty quickly
is there a logic, which branch is in front of another?
Pretty picture to illustrate an interesting conjecture. I bet there are a lot of interesting ways to illustrate the process.
This one was a treat
Is anyone else really bothered by the structure of those tree examples? Like at 3:43 the number 3 is connected to 5 when it should be connected to 10. 5 is connected directly to 8 when it should have to go through 16 first.
That isn't on purpose right? Maybe it was just explained confusingly??
It's in reverse order because the conjecture is that all branches die in 1, rather than start at 1 (which would just cycle 1-4-2-1).
Mage of Void is right imo. Connecting 5 to 8 makes as much sense as connecting 8 to 2: it's skipping a step. Sure, 5 will go to 8 as 8 will go to 2 but 5 gets to 8 through 16. Same with 13 to 20. It goes through 40 so should connect to it.
The diagram is consistent in combining an odd step with its subsequent even step so no technically wrong as a schematic but I don't see how that is helpful.
Mage of Void... the tree splits at those nodes. Even to the right, odd to the left.
...I see what you're saying now. The line should go through the bigger number only.
*3×3+1=10 , 10÷2=5*
N.J.Wildberger uses the Collatz conjecture to puncture ones naive notions of just how good infinite processes and definitions of convergence are. He says in analysis you only deal with easy sequences of numbers.
Can you make a link to the image please ?
If n is odd, 3n+1 is even, so one will immediately divided by 2, so one could make the step (3n+1)/1 before checking polarity again. The tree would then be straighter with know evens removed.
Can someone explain why you have e.g. 5->8 (3:44) when it really is 5->16->8?
When you have an odd number and multiply it by three, you will have an odd number. Add one and you get an even number. So then you divide it by two. For convenience, some people make (3n+1)/2 into one step. 5 goes directly to 8 instead of 16.
Beautiful video, thanks.
However, 3:07 the trees are a bit wrong if you follow the collatz conjecture rule, right?
3 should connect to 10 (and not to 5)
13 to 40 (and not to 20)
21 to 64 (not to 32)
85 to 256 (not 128)
Hmmm, looks like it is mentioned in 4:24 (for odd number do 3n+1 and then divide by 2). I am wondering, why didn't the artist use the original collatz conjecture rules but slightly altered them this way? To make the tree look more "compact" with branches closer together, or?
Thanks.
That was very interesting and beautiful
AWESOME !!
dude I want a shirt with that. somebody create one (or more) please, high quality.
That looks like my hair in the mornings
Are you a Medusa? O_o
yoou mean a gorgon. Medusa is an individual.
Mebbe
I mean, your avatar is a Trapinch, which has Arena Trap, so just like Medusa, if you look at it, you can't get away :3
You need some apply some collagen?
Get it?... Because, collagen... Collatz
All you have to do is prove that with any odd number, when following the steps, you eventually get to a power of 2. Idk if that's hard to do but it's a start if they don't have one.
I have some information to spread (ifn't it already exists) So first reverse the Conjecture so /2 is *2 and *3+1 is -1/3.
Then with those equations you can apply either *2 or -1/3 to even numbers, and you can only *2 odd numbers.
now if you do a tree with only 2^x you quickly realise that only x in this case must be even for it to be a choice between *2 or -1/3. And another cool thing is that if you add your current even 2^x with it's -1/3 result you'll get the next in the lines result.
Illustration:
1 4+1=5 16+5=21
/\ /\ /\
4 -> 8 -> 16 -> 32 -> 64 . . .
The equation for that being 2^x + (2^x-1)/3 = (2^(x+2)-1)/3.
I hope I provided something interesting.
Oh and btw *3+1 is related to *2+2 which is much easier to understand.
The picture is so satisfying!!!
nice job colouring perfectly between the lines
I didn't expect to see a speed drawing on that channel, you keep surprising me x)
I wanna color in that. Anyone got a source to the uncolored drawing?
If a n = 2^m, then applying the algorithm to n WILL give a sequence that achieves 1 at the iteration m. If n is even & NOT an integer power of 2, then there exists some number of iterations p such that after p iterations, the resulting number is an odd number. Therefore, the question we are interested in is equivalent to the question, "if n = 2m + 1, then can we always reach some power of 2 by having 3n + 1?" 3(2m + 1) + 1 = 6m + 4, and this is a power of 2, then 3m + 2 is a power of 2. We want 3p + 2 = 2^q with natural solutions p & q. If a starting number can achieve a solution p to this equation, then that number achieves 1. This is truly a question about modular arithmetic.
Awesome. Just awesome.
City and Colour - Bring me your love
album cover has that design on it. Pretty neato
That is amazing! I did my thesis making a model of seaweed (Ulva lactuca) growth. I love how something organic-looking can be "created by" such a short mathematical formula... And I'd totally love to have this as a poster on my wall. Guess I'll have to get the coloring book and start getting creative ^^'
Not "created by", just "described by". Mathematics is mankind's way of giving linguistic adjectives to nature, describing things and events in the abstract and extrapolating outwards to gain ever-more precise logical understanding of those things and events. I'm sure you already knew that though, since you did a thesis on it. I just wanted to make it clear for others who might read this and mistakenly think that mathematics creates things. If people like Stephen Hawking can make such a silly blunder, then anybody can.
Yeah thanks! That's why I put the "created by" in quotation marks :)
That's what I figured. Like I said, just wanted to clarify it for others.
Things like this, and most certainly the prime numbers, are concepts people will never understand; the intricacies of the positive integers transcend human thought. As Leopold Kronecker was quoted as saying, ""God made the integers, all else is the work of man."
Oh my gosh this is crazy, Edmund Harriss was my math professor last semester. This is so weird, but cool!
Are negative numbers not allowed? They have a different loop. -1 to -2 to -1. Also -5 -14 -7 -20 -10 -5 is its own loop.
Was there a systematic way that you chose which chains/branches/tentacles should be drawn on top of the others?
mvmlego1212 I have not checked it and i have no idea how they did it but i would guess by the value of each branch where they intersect. could just be random though.
Since a different person colored it, it's kinda hard to know.
It's something wrong with this image. The Collatz three branches off at all numbers of the form 6n+4 and only those. I call them nodes. A node creates a vertical branch (2n-branch) and a horisontal branch ( (n-1)/3-branch). This means that the nodes are 10, 16, 22, 28 and so on. It is simple to show that all numbers will end up in a node. It is also simple to show that all the odd numbers are located as the first number on a horisontal branch (and hence there is a one-to-one relationship between all odd numbers and all the nodes, and hence all the horisontal branches) (If you can show that all the nodes are connected you have hit the jack-pot...) There are three types of nodes, the ones of type 18k+4 (this is type 1), 18k+10 (this is type 2) and 18k+16 (node type 3). Nodes of type 2 creates a branch that does not have any new nodes. Type 2 nodes are 10, 28, 46 and so on. Only type 1 and 3 creates horisontal branches with new nodes. And there is a pattern to what type of node the next node is - it is not "chaotic" - even if it looks chaotic...
The long branch at the bottom of the figure does not have any new branches branching off - this means that this is a type 2 horisontal branch. This branch must start with an odd number, all the rest are even. One such horisontal branch of type 2 is 3,6,12,24,48, ..., 3*2^p - this branch originates from 10 (the node), but there are no new nodes on this brach - no number in this branch will ever equal 6n+4 and all of them except the first number are even (as are the case with all horisontal branches). This means that this branch in the figure shall always turn clockwise. But it clearly doesn't - it turns a little bit in this and a little bit in that direction..???... Actually it turns mostly anti-clockwise, which means that there are mostly odd numbers in the branch, which is impossible, since branches which do not have any new branches only have even numers in them - the numbers are e*2^p, where e is an even number in the series 3, 9, 15, 21, ... (the starting number of the branch). So 3, 6, 12, 24 is one such branch and 9, 18, 36, 72 is the next and 15, 30, 60, 120 the third type 2 branch and so on.
Hence the computer code that makes this figure has some type of "randomness" written into the code - making it look more chaotic than it really is. I hope that they can remove this randomness and show a "pure" one - I like playing with the collatz conjecture and would love to see it.
By the way, type 1 and type 3 nodes creates new nodes, and there is a simple pattern to what the new node is and what type of node the next one is. But showing that all the nodes are "connected" is the difficulty.....
Just thinking if it would be possible to calculate fractal dimension of this figure.
What would the tree look like if you only curved it when the parity changes and kept it straight otherwise?
odd has to be preceded by even, and followed by even. So for each 3n+1 there is a divide by 4, 2 before 2 after. So 18 > 9 > 28 > 14 It has to reduce to one. what am I missing? if you repeatedly multiply by 3 but divide by 4 I'm guessing it'll get smaller. and smallest positive integer is 1, or maybe zero?
*There is a sequence of numbers: 4 --> 2 --> **-40 --**> **-230 --**> **-748 --**> **-1846 --**> ?*
*What is the seventh number ?*
I think that look even more awesome if someone will print this in 3D!
congrats for 2Million subs! Any plans?
Why does it have small figures like sucker of an octopus under lines? It leads thinking of underwater creatures. And why do branches from even numbers overlap the lines from odd numbers?
I have a solution. Since every odd will have 1 added to it every odd will become even next turn. The only way to get down to 1 is from a number with a root of 2 like 2^3 or 2^n. So therefore, every odd number will become even and then odd again and over and over until the even is a root of 2.
For example, 13 goes to 40 which goes to 20, then 10, 5,16, 8, 4, 2, 1.
How do you know _EVERY_ starting number will go from even to odd until it finds a power of two?
OrangeCreeper217
Because every odd number will have 1 added to making that odd number even.
OmegaMike 108
Thank you for the oh so insightful commentary on how the conjecture works. ( ͡° ͜ʖ ͡°)
Seriously, though, not all even numbers are a power of 2. How would you prove that it goes to a power of 2 at one point? How do you know if there isn't a number that doesn't go to a power of 2 at all?
OrangeCreeper217
Beats me, I didnt think I would get this far.
those trees remaind me of philogenetic trees that show the common ancestor of several species. Wonder if we could ever apply this conjecture to predict patterns of evolutionary process, or something like that.
sorry for bad english
Does the Collatz Conjecture work if you instead use 5n+1, 7n+1,9n+1,etc? Also, is there a formal proof for n+1?
What would happen if you add # of iterations as a third component lets say going up and duration as in how long since each iteration was calculated as thickness of the branches? Would it look like a tree with the absence of wind and a perpendicular constant sun?
Beautiful video!
cheers
+Numberphile a = (pi*7) / (pi^7) Deserves discussion as it indicates that previous used constants to figure a are inaccurate?
I love that the tree is described as looking organic, so the artist colors it in with reds and pinks to make it look like meat.
or coral ...
So what he's saying is that there's no visible pattern? You could say it seems arbitrary for all we know.
But isn't the premise/goal in itself arbitrary and the result to be expected or at least not surprising?
Opposite conclusion depending on sign. odd->3*odd-1 versus odd->3*odd+1. So there is something in it.
The formula for the collatz conjecture is
N^2 + n - (2n+1)floor(n/2)
if you make all numbers even and divide by 2 why would you not always come back to one? i dont understand the mystique
How can you hold a pen like that and actually colour something? It looks like such a tense grip, it must be super tiring.
I think that is similar to proving that you can't win in a casino if you play long enough. The probability that the next number in sequence is odd is lower than the probability that it's going to be even. I imagine that the sequence length is exponentially smaller than the starting number. But I'm probably wrong :)
You will eventually reduce to the Collatz sequence (4,2,1) if you also use the following formula > (If EVEN n/2 if ODD 2n +2)
"Every number less than 10,000"
Is every branch extended until it reaches the first five-digit number? Or is every branch explored until every four-digit number has been found? I'm guessing he meant the former, because I once mapped out a tree for *every* two-digit number, and there were a couple of outlier branches significantly longer than the rest.
I don't see any long outliers in that image, which I would expect would have many more dramatic examples if including all four-digit numbers.
at 3:07, 5 should be linked to 16 not 8 right?
i have discover something amazing that like in abstartc algebra we have feilds like natural number is an subset of whol number and whole number is a subst of integers etc then in numbers like you see that prime and composite number is a subset of even and odd numer
I dont get it, i dont see anything about the collatz conjecture on that tree, help :(
Would this scale up? Like if you did it to every number above 10 million would it have more or less the same shape just with many more branches?
Can it be explained, why the tree is in majority on the right side instead of the left side?
Because as soon as you hit an odd number you make it even, so in the best case scenario for odd numbers 50% of them would be odd and 50% even.
This is obviously not the case as when you hit an even you can get either an even number or an odd one next.
The same reason that rivers flow downhill and time moves ever onwards: because the direction of a thing is built into the system by design.
I'd like a link to the specific song that was used for the coloring segments.
Hey can you guys do a video on TREE(3)? I'd like to learn more about its origins, the TREE function itself, and how much larger it is than Graham's Number because I know its quite a significant gap. Thank you!!
Would it be worth much if you could prove - definitively - that this could never blow up toward infinity? Or has that bit already been proven?