Why Busy Beaver hunters fear the Antihydra

https://benbrubaker.com/why-busy-beaver-hunters-fear-the-antihydra/

by Bogdanp • 9 months ago

238 78 9 months ago

russdill• 9 months ago

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.

gbacon3• 9 months ago

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)

hyghjiyhu• 9 months ago

Iirc if you change the numerical values of the collatz problem some instances are undecidable.

gbacon3• 9 months ago

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

_alternator_6• 9 months ago

Ok, I read this post quite a while ago and something about the reasoning bothered me then, and it still bothers me now.

In short, my read is that the argument does not rule out that there is a computable function that grows faster than BB(N), but rather it shows that it is impossible to prove or “decide” whether a given computable function grows faster than BB(N).

Maybe this is equivalent to the conclusion stated? Am I missing something obvious? (That sees likely; Scott Aaronson is much better at this than me.)

Edited for clarity

pmb• 9 months ago

Any computable function f on one variable x has a program. That function is a program of size p. The input x also has a data size d. BB(p+d) >= f(x), by definition, for all f and x. If you think you might have a (function, input) pair (and corresponding (program, data) pair) for which this is not true, see the previous sentence.

gbacon3• 9 months ago

Aaronson’s argument shows by contradiction that a computable upper bound D(N) that grows more rapidly than BB(N) cannot exist because otherwise we’d be able to use it to solve the halting problem.

KalMann4• 9 months ago

I think Scott's reasoning is correct in the end. If you suppose you had a computable function f(N) such that f(N) is always greater than BB(N). Then you could exploit the function f to solve the halting problem. Given a program of length N, run the program for f(N) steps. If it halts within that time, you know it's a halting program. If it doesn't halt within that time you know it will never halt.

jcranmer• 9 months ago

It is definitely way too easy to accidentally confuse "x is true" and "x is proven to be true" on topics of undecidability, and this is one of those times where the answer is in fact obvious, but it takes you 15 minutes to realize that it's obvious.

The halting problem is uncomputable (by diagonalization)--that is no computable function can solve the halting problem, rather than the weaker statement that we cannot assert that "this" function solves it. As a result, if somebody asserts a function that purports to solve the halting problem, then we can definitively say that either it isn't computable, or the thing it solves isn't the halting problem.

For the Busy Beaver function, it's obvious that the resulting program (if it existed) would solve the halting problem, so clearly it's not a computable function, and analyzing what part of it isn't computable leads you to the Busy Beaver as the only option.

For the D(N) function... well, since we assume D(N) ≥ BB(N), D(N) is still an upper bound on the number of steps a halting TM could run, so the resulting program using D(N) in lieu of BB(N) would still solve the halting problem, which forces us to conclude that it's not computable.

A different argument that may make more sense is this:

Consider the program "if machine M has not halted after f(sizeof M) steps, print not halt, else print halt." If f is a computable function, then the program is clearly computable. But since no computable program can solve the halting problem, we know that this program cannot either. Therefore, for every computable function f, there must exist some machine M such that M halts only after more than f(sizeof M) steps. In other words, f cannot be an upper bound on the Busy Beaver function.

kibwen• 9 months ago

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;
    }

OEIS link: https://oeis.org/A386909

EDIT 2: Of course the article mentions this in the next paragraph, which is what I get for being immediately nerd-sniped.

ameliaquining2• 9 months ago

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.)

achierius• 9 months ago

What else comes to mind in terms of inefficiencies? I can think of quite a few but you seem to have deeper insight as to their ranking so I'm curious as to your thoughts.

tgv• 9 months ago

Nitpicking: just as TMs can use binary or decimal arithmetic, so could Rust programs use unary. It's not an inefficiency in TMs per se, but I can see how it would help a TM to become a BB champion.

fijiaarone• 9 months ago

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.

ameliaquining2• 9 months ago

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.

weregiraffe• 9 months ago

sportsball

People who use this word should be banned from the internet for life.

MarcelOlsz• 9 months ago

If it's not loading for anyone else [0]

[0] https://web.archive.org/web/20251027173129/https://benbrubak...

CobrastanJorji• 9 months ago

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!

altruios2• 9 months ago

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?

wat100002• 9 months ago

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?

altruios2• 9 months ago

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.

A_D_E_P_T• 9 months ago

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.

LegionMammal9783• 9 months ago

The peak run lengths of evens/odds 'should' go to infinity, but these runs become a smaller and smaller component of the overall average, so that it is expected to approach the long-term 50% regardless.

In other words, an unbiased random walk should almost surely return to the origin, but a biased random walk will fail to return to the origin with nonzero probability. This can be considered a biased random walk [0], since the halting condition linearly moves further and further away from the expected value of the 50/50 walk.

[0] https://wiki.bbchallenge.org/wiki/Antihydra#Trajectory

emtel2• 9 months ago

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.

baruchel• 9 months ago

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.

henry2023• 9 months ago

One of the curiosities about this function is that computing BB(748) is independent of ZFC.

https://scottaaronson.blog/?p=4916

Sniffnoy2• 9 months ago

The record's been lowered since then, I should note. At least down into the 600s, I've seen claims of down in the 400s. But I haven't really kept up with this.

red75prime• 9 months ago

In other words, Turing machine creates mathematical reality, which is independent from our assumptions.

tromp3• 9 months ago

The BB function does grow mind bendingly fast. The machine running for 2↑↑2↑↑2↑↑9 steps is one of the 4^12*23836540 = 399910780272640 differently behaving 6-state machines [1].

A similarly fast growing function is the functional busy beaver [3]. Among all 77519927606 closed lambda terms of size <= 49 bits, there is one whose normal form size exceeds the vastly larger Graham's Number [3].

Several beaver fans believe that BB(7) might exceed Graham's Number as well, which struck me as unlikely enough to offer a $1k bet against it, the outcome of which will be known in under a decade.

[1] https://oeis.org/A107668

[2] https://oeis.org/A333479

[3] https://en.wikipedia.org/wiki/Graham%27s_number

Eliezer• 9 months ago

It would not surprise me at all for bb7 to exceed Graham's number. Just a Kirby-Paris hydra or a Goodstein sequence gets you to epsilon zero in the fast-growing hierarchy, where Graham is around omega+2.

JulianChastain2• 9 months ago

It seems totally inconceivable to me that you could accurately predict how long until we will know whether BB(7) is greater than Graham's Number.

throwaway815233• 9 months ago

best current lower bound

Well is there a best current UPPER bound, or at least a "probvious" one?

cevi• 9 months ago

There is no general procedure for computing upper bounds on busy beaver numbers (this can be proven). We haven't even come close to enumerating all of the interesting six-state Turing machines, so right now we don't even have a wild guess for an upper bound on BB(6).

shawnz2• 9 months ago

I think finding an upper bound is basically just as difficult as finding the actual value itself, since both would require proving that all of the programs which run longer than that will run forever. That's why we can say BB(x) grows faster than any computable function. Being able to compute BB(x) algorithmically or any faster growing function would let you solve the halting problem

chtl• 9 months ago

There is of course not currently any upper bound on BB(6).

I find the question about a probvious upper bound more interesting - There is also not a probvious upper bound on BB(6), as this would require at least some understanding of the high-level behavior of all remaining holdout machines. However, there may soon be a 'probvious' upper bound on the value of BB(3,3) (BB for 3-state, 3-symbol machines). Up to equivalence, there are four remaining machines to decide to find the value of BB(3,3). One is a 'probviously halting' machine which will be the new champion if it halts, and for which probabilistic models of its behavior predict with high probability an exact halting time. One is a 'probviously nonhalting' machine. The two other machines are not well-understood enough to say whether they have any probvious long term behavior, but some suspect that they both 'probviously nonhalt'. If this is true it could be said that a 'probvious' upper bound exists for BB(3,3).

entaloneralie5• 9 months ago

Immediately implements the Antihydra in Fractran

13122 -> 50/18, 55/42, 539/2, 297/275, 2/55, 2/11

73737373734• 9 months ago

This is cool, could you please explain your implementation?

entaloneralie5• 9 months ago

For sure, here are the registers used for each fraction:

  comp h^2 > comp H^2
  comp h c > done H
  comp     > done c^2
  done H^2 > done h^3
  done H   > comp
  done     > comp

  Accumulator: comp h^8

The first two does the comparison for odd/even, the h register is moved to the H register during the comparison, then done does H += H>>2, and then keep trying.

https://wiki.xxiivv.com/site/fractran.html

sligocki4• 9 months ago

50/18 reduces to 25/9 right?

sligocki4• 9 months ago

And 297/275 to 27/25?

entaloneralie5• 9 months ago

Conway's Fractran traditionally compares the accumulator against reduced fractions, but computationally-speaking, getting to the gcd of a fraction does little more than getting rid of otherwise valuable information used during comparison to match against a restricted set of fractions. The support for catalysts, symbols found on both sides of a rewrite rule, makes for a simpler and faster implementation.

  15/6 red [green] > blue [green]
  5/2 red > blue

  red green

These two fractions are not equal and reducing the first into the second, eliminates the capability to match against it only when the catalyst green is present in the accumulator.

Why Busy Beaver hunters fear the Antihydra

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