The best current lower bound for BB(6) is 2↑↑2↑↑2↑↑9 (google "Knuth Up Arrow" if this makes no sense), a number so inconceivably large it gives me the willies.
In particular, this means that in going from BB(5) to BB(6), you have already crossed the line where the actual busy beaver TM can no longer be simulated step by step in the lifetime of our universe (or a googol lifetimes of our universe for that matter).
It really is mind bending how fast this function grows.
> It really is mind bending how fast this function grows.
While the BB function is obviously a well-defined function over the integers, I find it helpful to think of it as a function over qualitatively heterogeneous items—such as stones, bread toasters, mechanical watches, and computers. The key idea is to view the underlying computing devices not as “a little more powerful” than the previous ones, but as fundamentally different kinds of entities.
TLDR; As BB(n) gets larger, they can encode more random walk style problems that have a stop condition related to the position of the random walk. Proving that such a condition is unlikely may be easy, but proving it never occurs is very difficult.
That was a really well written article. I think even somebody who had never heard of a Turing Machine could probably have gotten a pretty reasonable quick understanding of roughly what BB(5) and BB(6) are and how the Antihydra works and its greater mathematical/historical context. That's hard to do, good job!
Why we care about Busy Beaver numbers, from “Who Can Name the Bigger Number?” by Scott Aaronson:
Now, suppose we knew the Nth Busy Beaver number, which we’ll call BB(N). Then we could decide whether any Turing machine with N rules halts on a blank tape. We’d just have to run the machine: if it halts, fine; but if it doesn’t halt within BB(N) steps, then we know it never will halt, since BB(N) is the maximum number of steps it could make before halting. Similarly, if you knew that all mortals died before age 200, then if Sally lived to be 200, you could conclude that Sally was immortal. So no Turing machine can list the Busy Beaver numbers—for if it could, it could solve the Halting Problem, which we already know is impossible.
But here’s a curious fact. Suppose we could name a number greater than the Nth Busy Beaver number BB(N). Call this number D for dam, since like a beaver dam, it’s a roof for the Busy Beaver below. With D in hand, computing BB(N) itself becomes easy: we just need to simulate all the Turing machines with N rules. The ones that haven’t halted within D steps—the ones that bash through the dam’s roof—never will halt. So we can list exactly which machines halt, and among these, the maximum number of steps that any machine takes before it halts is BB(N).
Conclusion? The sequence of Busy Beaver numbers, BB(1), BB(2), and so on, grows faster than any computable sequence. Faster than exponentials, stacked exponentials, the Ackermann sequence, you name it. Because if a Turing machine could compute a sequence that grows faster than Busy Beaver, then it could use that sequence to obtain the D‘s—the beaver dams. And with those D’s, it could list the Busy Beaver numbers, which (sound familiar?) we already know is impossible. The Busy Beaver sequence is non-computable, solely because it grows stupendously fast—too fast for any computer to keep up with it, even in principle.
Despite all the reporting on BB(5) I had never seen anyone convey that equivalent high-level formulation from 1993, that's very cool!
EDIT: For fun I converted it to Rust and expected to see it spew a few million numbers into my terminal, but no, this equivalent loop actually terminates after 15 steps, which is fascinating given that the Turing machine takes 47 million steps:
let mut x = 0;
loop {
x = match x % 3 {
0 => 5 * x + 18,
1 => 5 * x + 22,
_ => break
} / 3;
}
This is because your Rust program represents the numbers in binary, while the BB(5) champion Turing machine represents them in unary. And unary arithmetic is exponentially slower than place-value arithmetic, which is why we invented the latter. (There are other inefficiencies in the Turing machine, but that's the big conceptual one.)
It is by definition not random. The Antihydra is generated by a fixed computable map, so it is compressible and would fail some effective statistical tests. You can't get true randomness via a deterministic algorithm; any computable infinite sequence fails Martin‑Löf randomness.
That said, empirically and in all current analyses, the Antihydra's parity behaves as if it were roughly fair over long spans (neither a proven odd nor even bias), and the short-range statistics look pseudo-random. Non-halting is overwhelmingly plausible... but a concrete proof seems out of reach.
I don't think a truly random sequence would necessarily halt under these rules. It's not enough to have arbitrarily long runs. As the sequence as a whole gets larger, the run length needed to end it also gets longer, and thus the probability gets smaller. The result should be something like a geometric sequence with a finite sum.
Consider a simpler version, where you flip a coin three times, then four times, then five times, etc., and you stop if you ever get the same side for every flip in a given turn. The probability that you'll stop is equal to 1/4 + 1/8 + 1/16 + ... which is 50%. If you do this forever then you'll eventually see a run of ten trillion heads or tails, but you probably won't see that run before your ten trillionth turn.
So I think the question is, does the anti-hydra sequence ever diverge sufficiently from randomness?
> As the sequence as a whole gets larger, the run length needed to end it also gets longer, and thus the probability gets smaller. The result should be something like a geometric sequence with a finite sum.
This is true.
But it would still halt. Infinity is weird like that.
To be clear, I mean the sequence of coin flips where the total value of heads/tails is 2:1.
The probability of having a 2:1 ratio of heads/tails - at some point - in an infinite sequence of fair flips is 1, is it not?
The anti-hydra may have a bias, and only if that bias is against the halt condition do we have a case where we can conclude that the anti-hydra does not halt.
No, it's not a distributed-computing thing; raw compute isn't the bottleneck. (That would only help if there were a need to check many machines that halt after a tractable-but-nontrivial number of steps; that's a narrow sweet spot, given the superexponential nature of the problem, and few machines of interest are believed to be in it.) Rather, it's a collaboration of human (mostly amateur) mathematicians chipping away at different parts of the problem.
The best current lower bound for BB(6) is 2↑↑2↑↑2↑↑9 (google "Knuth Up Arrow" if this makes no sense), a number so inconceivably large it gives me the willies.
In particular, this means that in going from BB(5) to BB(6), you have already crossed the line where the actual busy beaver TM can no longer be simulated step by step in the lifetime of our universe (or a googol lifetimes of our universe for that matter).
It really is mind bending how fast this function grows.
> It really is mind bending how fast this function grows.
While the BB function is obviously a well-defined function over the integers, I find it helpful to think of it as a function over qualitatively heterogeneous items—such as stones, bread toasters, mechanical watches, and computers. The key idea is to view the underlying computing devices not as “a little more powerful” than the previous ones, but as fundamentally different kinds of entities.
TLDR; As BB(n) gets larger, they can encode more random walk style problems that have a stop condition related to the position of the random walk. Proving that such a condition is unlikely may be easy, but proving it never occurs is very difficult.
Way worse than just very difficult:
- “Avoid the Collatz Conjecture at All Costs!” (Math Kook) https://www.youtube.com/watch?v=TxBRcwkRjmc
- “Experienced mathematicians warn up-and-comers to stay away from the Collatz conjecture. It’s a siren song, they say: Fall under its trance and you may never do meaningful work again.” https://www.quantamagazine.org/mathematician-proves-huge-res...
- "Mathematics is not yet ready for such problems [as Collatz].” (Paul Erdős)
Iirc if you change the numerical values of the collatz problem some instances are undecidable.
A generalized Collatz problem ((mx + b mod n) instead of 3x+1 in Z) is undecidable.
That was a really well written article. I think even somebody who had never heard of a Turing Machine could probably have gotten a pretty reasonable quick understanding of roughly what BB(5) and BB(6) are and how the Antihydra works and its greater mathematical/historical context. That's hard to do, good job!
If it's not loading for anyone else [0]
[0] https://web.archive.org/web/20251027173129/https://benbrubak...
Why we care about Busy Beaver numbers, from “Who Can Name the Bigger Number?” by Scott Aaronson:
Now, suppose we knew the Nth Busy Beaver number, which we’ll call BB(N). Then we could decide whether any Turing machine with N rules halts on a blank tape. We’d just have to run the machine: if it halts, fine; but if it doesn’t halt within BB(N) steps, then we know it never will halt, since BB(N) is the maximum number of steps it could make before halting. Similarly, if you knew that all mortals died before age 200, then if Sally lived to be 200, you could conclude that Sally was immortal. So no Turing machine can list the Busy Beaver numbers—for if it could, it could solve the Halting Problem, which we already know is impossible.
But here’s a curious fact. Suppose we could name a number greater than the Nth Busy Beaver number BB(N). Call this number D for dam, since like a beaver dam, it’s a roof for the Busy Beaver below. With D in hand, computing BB(N) itself becomes easy: we just need to simulate all the Turing machines with N rules. The ones that haven’t halted within D steps—the ones that bash through the dam’s roof—never will halt. So we can list exactly which machines halt, and among these, the maximum number of steps that any machine takes before it halts is BB(N).
Conclusion? The sequence of Busy Beaver numbers, BB(1), BB(2), and so on, grows faster than any computable sequence. Faster than exponentials, stacked exponentials, the Ackermann sequence, you name it. Because if a Turing machine could compute a sequence that grows faster than Busy Beaver, then it could use that sequence to obtain the D‘s—the beaver dams. And with those D’s, it could list the Busy Beaver numbers, which (sound familiar?) we already know is impossible. The Busy Beaver sequence is non-computable, solely because it grows stupendously fast—too fast for any computer to keep up with it, even in principle.
https://www.scottaaronson.com/writings/bignumbers.html
Despite all the reporting on BB(5) I had never seen anyone convey that equivalent high-level formulation from 1993, that's very cool!
EDIT: For fun I converted it to Rust and expected to see it spew a few million numbers into my terminal, but no, this equivalent loop actually terminates after 15 steps, which is fascinating given that the Turing machine takes 47 million steps:
OEIS link: https://oeis.org/A386909EDIT 2: Of course the article mentions this in the next paragraph, which is what I get for being immediately nerd-sniped.
This is because your Rust program represents the numbers in binary, while the BB(5) champion Turing machine represents them in unary. And unary arithmetic is exponentially slower than place-value arithmetic, which is why we invented the latter. (There are other inefficiencies in the Turing machine, but that's the big conceptual one.)
The Antihydra will halt if:
The sequence is (truly/fairly) random in its distribution of mods 1/2.
Even fair coins flipped infinitely would - on occasion - have arbitrary long results of heads or tails.
So the question becomes, is the anti-hydra sequence 'sufficiently' random?
It is by definition not random. The Antihydra is generated by a fixed computable map, so it is compressible and would fail some effective statistical tests. You can't get true randomness via a deterministic algorithm; any computable infinite sequence fails Martin‑Löf randomness.
That said, empirically and in all current analyses, the Antihydra's parity behaves as if it were roughly fair over long spans (neither a proven odd nor even bias), and the short-range statistics look pseudo-random. Non-halting is overwhelmingly plausible... but a concrete proof seems out of reach.
I don't think a truly random sequence would necessarily halt under these rules. It's not enough to have arbitrarily long runs. As the sequence as a whole gets larger, the run length needed to end it also gets longer, and thus the probability gets smaller. The result should be something like a geometric sequence with a finite sum.
Consider a simpler version, where you flip a coin three times, then four times, then five times, etc., and you stop if you ever get the same side for every flip in a given turn. The probability that you'll stop is equal to 1/4 + 1/8 + 1/16 + ... which is 50%. If you do this forever then you'll eventually see a run of ten trillion heads or tails, but you probably won't see that run before your ten trillionth turn.
So I think the question is, does the anti-hydra sequence ever diverge sufficiently from randomness?
> As the sequence as a whole gets larger, the run length needed to end it also gets longer, and thus the probability gets smaller. The result should be something like a geometric sequence with a finite sum.
This is true.
But it would still halt. Infinity is weird like that. To be clear, I mean the sequence of coin flips where the total value of heads/tails is 2:1.
The probability of having a 2:1 ratio of heads/tails - at some point - in an infinite sequence of fair flips is 1, is it not?
The anti-hydra may have a bias, and only if that bias is against the halt condition do we have a case where we can conclude that the anti-hydra does not halt.
> But it would still halt. Infinity is weird like that
What are you tring to say?
> The probability of having a 2:1 ratio of heads/tails - at some point - in an infinite sequence of fair flips is 1, is it not?
Yes, but "probability = 1" absolutely does not mean "will happen eventually" in pure mathematics. Infinity is weird like that.
Is this like SETI@home, Bitcoin, and Ai code generation?
In the old days we used to just chop wood, and burn it to keep warm. Then sit down and watch the sportsball game on TV to waste time.
No, it's not a distributed-computing thing; raw compute isn't the bottleneck. (That would only help if there were a need to check many machines that halt after a tractable-but-nontrivial number of steps; that's a narrow sweet spot, given the superexponential nature of the problem, and few machines of interest are believed to be in it.) Rather, it's a collaboration of human (mostly amateur) mathematicians chipping away at different parts of the problem.