BB(10) > Graham's Number(github.com) |
BB(10) > Graham's Number(github.com) |
Consider TREE(3) -> https://www.youtube.com/watch?v=3P6DWAwwViU
Graham's Number is notable for being one of the first, being an actual relevant bound in an unrelated proof (not just the output of a function), of teaching people the Up Arrow Notation.
To me, BB(10), TREE(3) etc. being so large is only minorly more impressive than GG(1) where I define I define GG(x) as "Graham's Number + X"
And this is a fine example of this.
I've been hearing stories of mathematicians who are developing a new found appreciation for LLMs by having them explain the other LLM's paper. I guess the paper has the necessary details but possibly not the best exposition.
I've had a lot of success answering questions like this by asking an LLM. It turns out they understand each other and do a great job explaining.
That said, functionally, they're more or less a restatement of the halting problem, essentially "if the halting problem was a sequence of numbers". Also see Godel's incompleteness. A bit mind-blowing that you have sequences of integers you can't "know" (i.e., compute or verify), but that's about it. There's no chance of this being useful in the system of mathematics we have today.
Graham's number is not a sequence, so it doesn't matter that the BB sequence grows faster than any computable sequence.
Would the following statement be correct?
"There is a turing machine that cannot be proven whether it will terminate or not. For any n we can construct a proof that it will run longer than n. But we cannot construct a proof that for any n it will run longer than n. Due to this particular turing machine, BB(k) cannot be determined for k large enough to admit that turing machine."
Every BB(n) is trivially computable, since any integer is computable. So for every BB(n), there is a computable sequence that grows faster than the BB numbers through n (annoyingly, given the word "grows", this can just be the constant sequence of BB(n) + 1).
So for any given computable number, the fact that it's computable gives no intrinsic reason to believe it's smaller than BB(n) for any particular n.
BB(10) > G > BB(5) = 47,176,870.Source: https://wiki.bbchallenge.org/wiki/Logical_independence
There is a point where these values become uncomputable, but they still exist (for some definition of exist).
Even for very small n you can create Turing machines where it's incredibly hard to prove that they halt: for n=25 you can create a Turing machine that stops only if the Goldbach conjecture is false.
Though that isn't enough to prove it's not computable, the point stands that if you could compute BB(n) for arbitrary n, you could run a Turing machine for BB(n)+1 steps and if it hasn't stopped yet then you know that the input corresponds to a machine that never halts. So it's impossible to write a program that computes BB(n) for arbitrary n, because doing so is equivalent to the halting problem.
And worse, there are some Turing machines where "proving the Turing machine stops" means "proving that a given set of axioms is consistent", where Gödel's incompleteness theorem says that it's impossible to build such a proof. So, that is how there is some n after which the busy beaver number is defined, but is not computable at all.
The busy beaver sequence is not computable (I am aware that somewhere around BB(700), you've encoded the consistency of ZFC). You cannot necessarily compute that BB(n) is a given number. But you can compute larger numbers. There is no such thing as integer that is too large to be computable.
I did slightly fuck up--the English language "every BB(n) is computable" is a bit ambiguous. What I should say is: "for all n, BB(n) is trivially computable, since it is a single integer".
It is not computable, because for high enough n you can't prove that you got the right integer.
And also you can prove that if a sequence a_n is always larger than BB(n) for some n, that sequence a_n is also not computable.