With fifth busy beaver, researchers approach computation's limits

https://www.quantamagazine.org/amateur-mathematicians-find-fifth-busy-beaver-turing-machine-20240702/

by LegionMammal978 • 2 years ago

550 137 2 years ago

quantamagazine.org

jl6• 2 years ago

BB(0) = 0

BB(1) = 1

BB(2) = 4

BB(3) = 6

BB(4) = 13

They just proved that BB(5) = 47,176,870.

It is known that BB(6) must be at least 10^10^...^10 (a tower of exponents fifteen levels high).

https://en.wikipedia.org/wiki/Busy_beaver#Known_values_for_%...

pdonis2• 2 years ago

The definitions of the "BB" function don't all appear to be the same. The article referenced in this overall discussion says BB(2) = 6. In the notation of the Wikipedia article you reference, this would be S(2) = 6; S is the number of steps. What has now been proved is that S(5) = 47,176,870.

However, your BB values for 0, 1, 2, 3, and 4 match the Wikipedia article's notation for Sigma; Sigma is the number of 1s written on the tape at halting. In that notation, what has now been proved is Sigma(5) = 4098.

LegionMammal97813• 2 years ago

Yeah, this is a bit confusing, and the subject of repeated internal controversy. Most of the twentieth-century authors focused on the number of 1s, Σ(n), following Radó's original practice of treating σ(M) as the score of a machine M in the "Busy Beaver game". But when Aaronson re-popularized it in 2020 [0], he used BB(n) to denote the number of steps (which Radó called S(n)), and the bbchallenge project has been using this latter convention for publicity. Pascal Michel's website [1] has all the Σ(n) and S(n) bounds up to n = 7.

Personally, I think both functions have their strengths and weaknesses. Σ(n) is easier to calculate for machines that run too long to be simulated directly (e.g., Skelet #1 from the article) but leave a known pattern on the tape, and it also has historical priority. But S(n) has a simpler argument for being undecidable, since it provides a trivial filter for testing if a candidate machine cannot halt. Also, σ(M) is a bit weird in that it has no lower bound in terms of s(M), since an adversarial machine could do a colossal amount of work before wiping its tape at the end.

Regardless, past BB(3), there isn't any known size where the champion machines for Σ(n) and S(n) are different. (At least, the sets of champion machines aren't disjoint: Σ(5) = 4098 is shared by both the S(5) champion and another machine that runs a quarter as long.) The score of a machine is dominated by googological strength rather than technicalities in the definition.

[0] https://scottaaronson.blog/?p=4916

[1] https://bbchallenge.org/~pascal.michel/ha

Natsu2• 2 years ago

Looking at the pictures on Wikipedia regarding those programs, it's interesting how they seem to result in fractals.

LegionMammal97813• 2 years ago

This mainly has to do with the BB(5) champion having "bouncing" behavior: it moves from the left side of the current pattern to the right side and back again, each time extending it slightly. Since the time it takes to grow the pattern is strictly proportional to its current size, you end up with a bunch of parabolas that appear self-similar.

jerf3• 2 years ago

That is because these are small. Small machines can not help but still have those sorts of patterns in them. See all the simple 1-D automata of interest, for instance: https://atlas.wolfram.com/01/01/

As you step up I am fairly confident you'll see the winners get noisier and noisier. They'll have some sort of pattern, just ones you won't be able to comprehend. For some suitable definition of "noisier" this might even be a provable claim.

lmm2• 2 years ago

Fractals or fractal-like things seem to be the main way you get "interesting" behaviour out of very simple rules. Procedurally generated art/content tends to use them a lot (look at e.g. demoscene programs).

kryptiskt• 2 years ago

One notable thing here is that the proof is a Coq proof. I wonder if it is the first significant proof that starts out implemented in a theorem prover, instead of being a known proof translated to such a system.

Note that there have been computer-assisted proofs before (four-color theorem, Kepler's conjecture), but those were not done in a formally verified setting until later.

LegionMammal97813• 2 years ago

As far as I am aware, all of the proofs and techniques for each machine were present before mxdys managed to get the whole theorem into Coq: the main problem was that the deciders and manual proofs were disorganized and somewhat suspect. The worst offenders here were Skelet #1, which needed a bespoke program to accelerate it to its ultimate pattern [0], and Skelet #17, which took Xu seven pages' worth of dense reasoning to prove non-halting [1]. The full Coq proof put a much-needed degree of confidence into these results.

[0] https://www.sligocki.com/2023/03/13/skelet-1-infinite.html

[1] https://discuss.bbchallenge.org/t/skelet-17-does-not-halt/18...

poikroequ• 2 years ago

[The four color theorem] was the first major theorem to be proved using a computer.

https://en.m.wikipedia.org/wiki/Four_color_theorem

I guess maybe I don't understand what you mean by "formally verified setting", but I believe the four color theorem was first proven using a computer.

Although flawed, Kempe's original purported proof of the four color theorem provided some of the basic tools later used to prove it.

It sounds like Kempe laid some of the groundwork, but then the theorem was ultimately proved using a computer.

I could be wrong though, I'm not an expert in this area.

nyssos• 2 years ago

The original four color theorem proof used a computer as a computational aid for some nasty casework: the procedure for checking each case and the list of cases that needed to be checked were found by hand.

Proving something in a theorem prover means the proof itself is an object constructed in the prover's language.

MaxBarraclough• 2 years ago

I'm no expert either, but it's interesting to note that the original computer-aided proof was erroneous. ctrl-f for Schmidt on the Wikipedia page.

ks20482• 2 years ago

Apparently, this is the proof in 19,000 lines of Coq: https://github.com/ccz181078/Coq-BB5/blob/main/BB52Theorem.v

nwallin• 2 years ago

Attempts to prove BB(5) would have begun long before theorem provers were around. This BB was discovered in 1990, and all machines of size 5 would have been enumerated not long after.

lupire5• 2 years ago

Enumerating machines is trivial. Understanding and analyzing their behavior is hard.

bryan0• 2 years ago

Aaronson’s post on BB(5): https://scottaaronson.blog/?p=8088

pvg• 2 years ago

Some comments on this result by Scott Aaronson https://scottaaronson.blog/?p=8088

And for leisure-class beavers, some big related threads from earlier this year:

https://news.ycombinator.com/item?id=40453221

https://news.ycombinator.com/item?id=38113792

https://news.ycombinator.com/item?id=37910297

inhumantsar• 2 years ago

leisure-class beavers

I can't not find this phrase funny. out of context it reads like something out of Terry Pratchett or Douglas Adams.

IIAOPSW• 2 years ago

frankly, leisure beaver sounds like a raunchy innuendo

max-ibel• 2 years ago

Exactly. Those are also not funny at all.

phaedrus• 2 years ago

I wrote program to solve the cutting stock problem (https://en.wikipedia.org/wiki/Cutting_stock_problem) for a personal project. I couldn't (or didn't want to) use any existing program for it because my stock involved cutting pieces shaped like either /---/, /---|, or |---| and I didn't want to waste material on the 45 cut.

I find it interesting that the description of Brady's program to optimize search for BB(4) by cutting out search subtrees whose differences don't matter is fairly close to a description of what I did to make my program fast.

wodenokoto2• 2 years ago

So we were just lucky that all non-halting programs of length 5 just happened to be provably non-halting?

tromp4• 2 years ago

The article touches on that:

But just four days ago, mxdys and another contributor known as Racheline discovered a barrier for BB(6) that seems insurmountable: a six-rule machine whose halting problem resembles a famously intractable math problem called the Collatz conjecture. Connections between Turing machines and the Collatz conjecture date back to a 1993 paper by the mathematician Pascal Michel, but the newly discovered machine, dubbed “Antihydra,” is the smallest one that appears unsolvable without a conceptual breakthrough in mathematics.

dwh452• 2 years ago

I'm curious how the Turing machines can resemble problems which take input? BB(n) is defined as a n-state Turing machine that starts off with an empty tape. Collatz(n) is how many steps are taken before it terminates when starting with input 'n'.

Does this mean a BB(6) machine which resembles Collatz is testing all possible values as part of it program and not part of anything on the tape (since the tape start out empty)?

srcreigh13• 2 years ago

Bit of a side note, but the article is a bit misleading here.

Collatz-TM connections have been investigated much earlier.

My fav is John Conway in 1972, who showed that the generalized Collatz problem is undecidable.[1]

[1] https://en.wikipedia.org/wiki/Collatz_conjecture#Undecidable...

LegionMammal97813• 2 years ago

Yes. In fact, Allen Brady feared as early as 1988 that there would be a totally intractable machine with 5 states [0]:

Prediction 5. It will never be proved that Σ(5) = 1,915 and S(5) = 2,358,064. (Or, if any larger lower bounds are ever found, the new values may be substituted into the prediction.)
Reason: Nature has probably embedded among the five-state holdout machines one or more problems as illusive as the Goldbach Conjecture. Or, in other terms, there will likely be nonstopping recursive patterns which are beyond our powers of recognition.

Luckily, this prediction did not come to pass, but only by a margin of one extra state.

[0] Allen Brady, "The Busy Beaver Game and the Meaning of Life", in Rolf Herken (ed.), The Universal Turing Machine: A Half-Century Survey, Oxford University Press, 1988, pp. 259–277. This chapter can also be found in the 2nd ed., Springer, 1995, pp. 237–254.

Sniffnoy8• 2 years ago

When you say "provably", do you mean that in the mathematical sense or in a practical sense?

For the practical sense, other commenters have already replied.

For the mathematical sense, I would say it would be pretty surprising to me if BB(5) had been undecidable -- 5 states and 2 symbols is just so few with which to try to encode undecidable behavior.

However, it's worth noting that as a consequence of the incompleteness theorem, there must be some n such that standard mathematics cannot prove the value of BB(n). And in recent years various people have been working on finding such n and seeing how low they can get the number. The current record[0] is 745.

I expect that record can be probably be brought lower (more easily than the record for computing BB can be brought higher!), but even so, that's a lot of distance between 5 (the highest we know) and 745 (the lowest we know to be unknowable).

[0]This is the current record both for ZFC and for PA, in case you're wondering what I mean by "standard mathematics"... so it at least ought to be possible to bring it down further for PA! I think so far nobody's found a way to do better for PA than for ZFC, even though, like, that has to be possible, right?

ko27• 2 years ago

Are there any interesting mathematical problems that can be encoded with 5 states and 2 symbols, which now we can prove using BB(5)?

srcreigh13• 2 years ago

BB(745) being independent of ZFC isn’t due to incompleteness theorems. It encodes a different theorem which is also independent of ZFC.

https://scottaaronson.blog/?p=2725

wodenokoto2• 2 years ago

So I know we can prove that we cannot solve the halting problem for all TM.

But for any given TM, can we decide whether it is provable or not? Or will we meet some that we will never know if we can solve or not?

6nf• 2 years ago

5 states and 2 symbols is just so few with which to try to encode undecidable behavior.

Would you be brave enough to say the same thing about BB(6) or BB(7)?

tromp4• 2 years ago

There are many variants of the original busy beaver problem, and some Busy Beaver Challenge contributors plan to keep working on these.

One such variant is a functional busy beaver defined in terms of the lambda calculus [1]. Since it measures program size in bits rather than states, it allows many more values to be determined (37 so far versus only 6 for TMs), and the gap between the largest known value and values beyond Graham's Number is a mere 13 program bits. A closely related variant [2] can be directly expressed in terms of Kolmogorov complexity, which Mikhail Andreev argues [3] is crucial for applications in Information Theory.

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

[2] https://oeis.org/A361211

[3] https://arxiv.org/pdf/1703.05170

passion__desire• 2 years ago

Isn't there another formulation of BB where we count shifts (from left to right, and from right to left) a TM makes instead of strings of contiguous 1s. I remember seeing a video about this definition.

Sniffnoy8• 2 years ago

The variant of BB discussed in the article is counting steps, not 1s. (I guess "shifts" is equivalent to steps, although that seems like a more roundabout way of specifying it.) Also, I don't think anyone counts strings of contiguous 1s, although of course one could define such a thing.

anandijain• 2 years ago

A bit unrelated but you posted oeis so hoping you know. In the article they mention there are 17 trillion possible 5-2 turing machines. I tried finding the sequence for this but couldn't.

I found this https://oeis.org/A141475, but it gives 27 trillion for 5.

tromp4• 2 years ago

While the official formula [1] is (4(n+1))^(2n) , if one ignores the symbol written and head movement for transitions to halt, then this becomes (4n+1)^(2n), which is ~ 17 trillion for n=5.

[1] https://oeis.org/A052200

titanomachy2• 2 years ago

I worked for a couple years with a formidable and incomprehensibly smart engineer who ascended the IC ranks faster than anyone I’ve seen at an elite tech company. He quit the job a few years ago, and when I asked him his plans he told me he was going to work on the busy beaver problem. I can’t help but wonder if he is mxdys, the pseudonymous contributor mentioned in the article who wrapped up the formal proof of BB(5). I’ll probably never know.

jebarker2• 2 years ago

If it were them would you be surprised that they wanted to remain anonymous?

someplaceguy2• 2 years ago

I'm sorry, I don't get it. What are you suggesting to be the motivation to remain anonymous in this case?

titanomachy2• 2 years ago

I didn’t know him that well, but could see him being the kind of person who wants to avoid attention. He was happy to spend some time explaining the problem to me and why it was interesting and difficult.

optimalsolver• 2 years ago

How does figuring out the halting behavior of ever larger Turing machines help humanity?

What's the payoff here?

I would rather someone of the intellectual calibre you describe work on problems more relevant to improving the world.

qayxc• 2 years ago

What's the payoff here?

Personal happiness? It's also more than short-sighted to assume that just because there is no immediately obvious overall benefit, that no such thing exists.

The techniques and insights developed and found during the pursuit of such fundamental and elusive problems can have profound side effects that may materialize only decades or even centuries later.

kevinventullo• 2 years ago

What’s the payoff of writing a poem?

Etheryte• 2 years ago

At the time when they were first explored, you could've easily said the same thing about subatomic particles, imaginary numbers and many other concepts of the like. Scientific advancement isn't always linear and the value of many discoveries is often only apparent after the fact, often after a considerable amount of time has passed.

karmakaze• 2 years ago

I would classify advancing the state of the art knowledge in math/computers/science to be a noble feat in itself that doesn't need any justification (especially to those who have done less).

utensil4778• 2 years ago

This is just the same extremely poorly thought out argument against any type of pure research.

Pursuing knowledge for knowledge's sake always results in improvement down the line, even if it's impossible to predict when and how.

Do you think Benjamin Franklin expected a payoff from the experiments that led to the discovery of electricity? Or Newton researching gravity? Much of our knowledge about the universe comes from researchers picking at a topic just for the hell of it.

I would rather someone of the intellectual calibre you describe work on problems more relevant to improving the world.

Impressive amount of entitlement here. Nobody owes you anything, and people are free to do anything they want. Not everyone wants to be a hero or great inventor, even if they have the skill for it. Nor is it some universal moral obligation for skilled people to spend their lives producing something profitable or beneficial to society.

Do you have hobbies? If so, you should feel bad because you're wasting your talents on useless unprofitable activities. Better go out and solve the world's problems.

dwaltrip• 2 years ago

I’m sure he cares very little what you think he should do.

danavar• 2 years ago

A message on LinkedIn or email never hurt!

ks20482• 2 years ago

According to Scott Aaronson's blog post on this, there are 16,679,880,978,201 5-state Turing machines. I wonder if we know what percentage of them halt?

Edit: number of TM for n states: (4n + 1)^(2n). Found this (for smaller n), which is the kind of analysis I was curious about: https://github.com/LukasKalbertodt/beaver

srcreigh13• 2 years ago

We must know the percentage which halt! I can’t find it in the bbchallenge.org site, but every machine is categorized.

LegionMammal97813• 2 years ago

Not necessarily. The seed database only contains TMs in tree normal form (TNF), which already contains several reductions. To convert the percentage of TNF halters to the percentage of total halters, you'd have to account for the multiplicity of each TNF machine, which is a bunch of fiddly combinatorics. Also, you'd have to add the machines that have no TNF form due to never writing a 1 to the tape.

nayuki• 2 years ago

From the article:

To distill the essence of the halting problem into a simpler form, Radó imagined sorting Turing machines into groups based on how many rules they had — one group for all one-rule Turing machines, another for all two-rule machines, and so on. Sure, that leaves infinitely many such groups, since a Turing machine can have any number of rules. But the number of distinct machines in every group is finite, since there are only so many possible combinations of rules.
With two rules, there are already over 6,000 distinct Turing machines to consider; that number swells to millions with three rules, and to billions with four.

I'm pretty sure the standard terminology is "states", not "rules". This deviation made it harder to understand.

Each state produces 2 transition rules, depending on the symbol at the tape head.

smokel• 2 years ago

The original Busy Beaver paper by Tibor Radó ("On Non-Computable Functions") is actually quite easy and fun to read.

For a modern version of the paper with some additional notes, see https://data.jigsaw.nl/Rado_1962_OnNonComputableFunctions_Re...

nickdrozd6• 2 years ago

Congratulations to the team! So the (blank tape) halting problem is solved for 5-state 2-color Turing machine programs. Has anyone tried applying these same techniques to the 2-state 4-color case? That seems like it would probably be tractable, although generally speaking colors are more powerful than states, so there might be some surprises there. (6-state 2-color and 2-state 5-color both seem intractable, perhaps even provably so.)

By the way, there is an extremely stupid but disturbingly widespread idea that humans are able to just intuit solutions to the halting problem, using the mind's eye or quantum mechanics in the brain or whatever. Needless to say, this did not factor into the proof.

tromp4• 2 years ago

the 2-color 4-state case?

You mean the 2-state 4-color case...

nickdrozd6• 2 years ago

Fixed, thanks

LegionMammal97813• 2 years ago

Has anyone tried applying these same techniques to the 2-color 4-state case?

I assume you mean the 4-color case. As I understand it, the deciders currently in use are sufficient to prove all the 2×4 holdouts non-halting. So the current champion gives us Σ(2,4) = 2,050 and S(2,4) = 3,932,964, barring some big errors in the decider design. The result just hasn't been organized in one place.

(6-state 2-color and 2-state 5-color both seem intractable, perhaps even provably so.)

Yes, 2×5 has the Hydra, and 6×2 has the Antihydra, which compute the same iteration, but with different starting points and halting conditions. The standard conjecture (related to Mahler's 3/2 problem) is that this iteration is uniformly distributed mod 2, and a proof of that conjecture would very likely prove both machines non-halting, by yielding suprema and infima on the cumulative ratio of 0s to 1s. But of course there is no known method of proof.

srcreigh13• 2 years ago

By the way, there is an extremely stupid but disturbingly widespread idea that humans are able to just intuit solutions to the halting problem, using the mind's eye or quantum mechanics in the brain or whatever. Needless to say, this did not factor into the proof.

The year is 52,000 CE and humans have solved BB(18) in the sense of exhaustively categorizing halting vs non-halting 19-state no-input programs. They have used a proof generator based on a logical theory called Aleph*, and at that time it had been known for 1.5k years that ZFC is incapable of establishing BB(18).

Compared to the year 2024 CE, considerable millennia before Aleph* came into use, it is clear that no program written at that point in history was capable of even using brute force proof checking to solve BB(18) in theory (like how we can enumerate and check ZFC proofs today to solve BB(??) in theory).

That's what is meant by the "humans intuit solutions to the halting problem" position. AFAIK, there's no known hard, theoretical reason why the above laid out future history cannot take place. And due to BB being incomputable, humans had to develop new theory to be able to construct the programs required. Something has to be accredited for the results, and it can't be computation since the programs did not exist.

LegionMammal97813• 2 years ago

AFAIK, there's no known hard, theoretical reason why the above laid out future history cannot take place.

Probably the biggest issue is that they'd have no method to establish that Aleph* is consistent. To continue this BB chain indefinitely, you must invent further and further first-order theories, each of which might not be consistent, let alone Σ₁-sound. And with an Σ₁-unsound theory, any halting proof might not hold up in the standard model of the integers. You'd effectively have to postulate an indefinite amount of oracular knowledge.

Also, another physical issue: you can show that within any consistent, recursively axiomatizable theory, the length of the shortest proof of "the longest-running n-state machine is M" must grow at an uncomputable rate in terms of n. Ditto for the shortest proof of "machine M halts", where M is factually the longest-running n-state machine. Otherwise, a machine could use a computable bound on the proof length to solve the halting problem. Therefore, the proof should very quickly become too large to fit within our light cone.

In any case, the BB-related evidence for that position rested on BB(5) being determinable by extending the techniques used for BB(4). But in fact, it turns out that similar extensions don't even get you to BB(6). So there isn't anything to support the position, other than the pure speculation that anything is physically achievable given enough time.

nickdrozd6• 2 years ago

They have used a proof generator based on a logical theory...

I don't understand your scenario. If they're using a proof generator, that sounds like the opposite of intuiting or using the human mind. Maybe they used "intuition" to come up with new axioms for a logical theory, but that is not the same as determining of some particular concrete TM program whether or not it halts.

lupire5• 2 years ago

Why would Aleph* be necessary or relevant?

ZFC and larger theories are only relevant for infinite objects, not finite objects like BB.

ganzuul• 2 years ago

extremely stupid

This is simply a test for if consciousness has infinite computational resources.

wayeq• 2 years ago

"After decades of uncertainty, a motley team of programmers has proved precisely how complicated simple computer programs can get."

my team has also proved this via our production codebase

kaptainscarlet• 2 years ago

These researchers clearly haven't looked at our NodeJS ERP system.

DowsingSpoon2• 2 years ago

Maybe I’m just an unsophisticated code monkey, but I read a little about Busy Beaver from time to time and I just don’t get it. Why is this an interesting problem? What do we hope to learn from it?

lkuty• 2 years ago

IMO, because it gives you a way to know if a program will terminate or not. If it consists of 5 states and goes beyond 47176870 steps of execution, then it will never halt. However, we are currently limited to 1 to 5 states Turing programs. And you can make correspondance between some Turing machines and some theorems in Mathematics thus it gives you a way to prove them I guess.

DowsingSpoon2• 2 years ago

Thank you! I’m trying to read up on it now also because this is definitely not something I know much about.

suzzer99• 2 years ago

The one thing I didn't understand is the distribution 1s and 0s on the tape. Does each Turing machine have to be solved for all possible combinations of 1s and 0s?

MrCheeze• 2 years ago

https://www.scottaaronson.com/writings/bignumbers.html

GTP2• 2 years ago

Why is this an interesting problem? What do we hope to learn from it?

Because it's a complicated puzzle, not necessarily because we hope to learn something from it. Which could happen, but it's definitely not the primary goal here.

seeknotfind• 2 years ago

Who else thinks mxdys is an AI program gone rogue? Thank you sweet mxdys!

Jean-Papoulos• 2 years ago

Can someone explain how there can a finite number of rules ? Could the rule "If it's a 0, change it to 1 and move 3 squares to the left" produce a different result than when moving 4 squares, or is that not the case ?

hmry• 2 years ago

You can only ever move one step to the left or right. If you want to move multiple steps, you need multiple states to use as a counter, or set up the tape a specific way ahead of time so you know when to stop moving.

masklinn• 2 years ago

A turing machine rule can only:

- set the current cell’s value

- move by one cell

- switch to the next rule (or halt)

(A TM rule has one sub-rule for each valid value / symbol)

Turing machines are characterised by the number of values (or symbols) and the number of rules (or states) (and technically the number of directions but TMs are generally one-dimensional). The Busy Beaver game fixes the number of symbols to 2, and only varies the number of states e.g. BB(3) is played on 3-states 2-symbols turing machines.

Thus the number of possible rules in BB(n) (n-states 2 symbols turing machine) is necessarily limited, to (4n + 4)^2n.

lupire5• 2 years ago

https://scottaaronson.blog/?p=8088

You need to find a mathematical reason why it can’t halt, and there’s no systematic method for finding such reasons—that was the great discovery of Gödel and Turing nearly a century ago.

That's only true for sufficiently large N, large enough to encode the halting paradox. (How large is that N? It's larger than 5!) Smaller N can and do fall to systematic methods.

(x) = (5x+18)/3 if x = 0 (mod 3),

The second = should be \equivalent. This is a rare case where that actually matters, because is being used in both non-modular integer operations (divide by 3) and modular operations (where division by 3 is not defined).

openasocket• 2 years ago

All things considered, the proof is pretty short! It’s 19,000 lines of Coq (including white space and comments). And in my experience, if this were compiled into a traditional paper it would be significantly shorter than the Coq version. (Of course, length of the proof really isn’t a measure of difficulty or complexity, but it is a very rough yardstick we can use.)

When talking about the limits of human knowledge, I often think about the theorems that are provable, but so complex that no human being could understand them. Probably the most complex proof we have is the classification of finite simple groups, which is thousands upon thousands of pages, and there is probably very few or no people on Earth that fully understand all of it.

As the article suggests, BB(6) could be undecidable. But it could also have a proof millions of pages long, beyond the reach of humanity.

With fifth busy beaver, researchers approach computation's limits

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