710: Collatz Conjecture

Explain xkcd: It's 'cause you're dumb.
Jump to: navigation, search
Collatz Conjecture
The Strong Collatz Conjecture states that this holds for any set of obsessively-hand-applied rules.
Title text: The Strong Collatz Conjecture states that this holds for any set of obsessively-hand-applied rules.

[edit] Explanation

The Collatz conjecture is a longstanding unsolved problem in mathematics. It states that repeating the sequence of operations described in the comic will eventually lead to the number 1. The description in the comic starts out accurate, then veers into the joke.

The comic illustrates the sequence with a graph in which an arrow connects each number to its successor. For example, the number 22 is even, so the next number in the sequence is 22 ÷ 2 = 11, and there is an arrow from 22 to 11. On the other hand 11 is odd, so the next number is 3 × 11 + 1 = 34, and there is an arrow from 11 to 34.

According to the title text, Cueball is obsessively writing out the graph by hand, and is so preoccupied with the task that he has stopped socializing with his friends. He will be busy for a very long time, because the Collatz conjecture has been confirmed for all starting values up to 5 × 1018.

The Strong Collatz Conjecture in the title text is a humorous extension of the Collatz Conjecture. Some other mathematical conjectures and axioms also have normal and Strong variants, where the Strong variant gives a more general rule. This practice is further parodied in 1310: Goldbach Conjectures.

[edit] Transcript

[Cueball sits in a chair at a desk, papers piled on top, writing furiously. Depicted above are apparently the writing, a series of nodes in various Collatz sequences (starting with 7, 21, 24, 29, 106, 176 and 256), all eventually leading back to 1.]
The Collatz Conjecture states that if you pick a number, and if it's even divide it by two and if it's odd multiply it by three and add one, and you repeat this procedure long enough, eventually your friends will stop calling to see if you want to hang out.
comment.png add a comment!


Technically, the 1 should lead to the four, causeing a loop. Because (1*3)+1=4, 4/2=2, 2/2=1 xD ‎ (talk) (please sign your comments with ~~~~)

I'm pretty sure (follow the Wiki link, perhaps, although I haven't yet) there's an implicit "...eventually lead to one, then you can stop" for this process.
And (again without checking the Wiki link), I suppose the conjecture could fail in one of two cases. Firstly if multiplying by three and adding one would take us to another odd number. Which cannot happen, because (odd*3) will be odd, so (odd*3)+1 is even. Which leads us to the possibility that the even number leads back to a prior odd number, to circle around again. There's no trivial case of (n*3)+1 => m, (m/2) => n, although there is the case of (n*3)+1 => m, (m/2) => o, (o/2) => n, for n=1. How, though, could we evaluate f1|2(xi) => xi+1 for all countably finite limits to i, given the rules of which f() function to use, to ensure that xi never equals x0. Now, that's a question and a half. Which I suspect has already been asked. And a half. 21:36, 7 May 2013 (UTC)
It seems way too general to be much more than "asked," and I am sure that it has been addressed in its simpler forms. In any case, there is enough amateur, recreational, and serious mathematical literature on it to find out that there are indeed two failure cases: a starting Collatz number results in an infinitely increasing sequence, or a loop exists apart from the 4-2-1 loop. (Curiously enough, some loops exist when negative numbers are allowed.) Stuff like this and Goldbach made me realize just how hard simple things can get. --Quicksilver (talk) 03:04, 20 August 2013 (UTC)
Personal tools


It seems you are using noscript, which is stopping our project wonderful ads from working. Explain xkcd uses ads to pay for bandwidth, and we manually approve all our advertisers, and our ads are restricted to unobtrusive images and slow animated GIFs. If you found this site helpful, please consider whitelisting us.

Want to advertise with us, or donate to us with Paypal or Bitcoin?