With fifth busy beaver, researchers approach computation's limits
quantamagazine.org
quantamagazine.org
And for leisure-class beavers, some big related threads from earlier this year:
https://news.ycombinator.com/item?id=40453221
I can't not find this phrase funny. out of context it reads like something out of Terry Pratchett or Douglas Adams.
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.
I found this https://oeis.org/A141475, but it gives 27 trillion for 5.
To me, that makes your conjecture very unlikely to be true...
What's the payoff here?
I would rather someone of the intellectual calibre you describe work on problems more relevant to improving the world.
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.
Why is it a smart person should be expected to use their capabilities for "the betterment of humanity"? There's a level of entitlement that goes with statements like that.
also the modern complex plane definition is not from Cardano.
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.
G.H. Hardy was a British mathematician who expressed pride in the "uselessness" of his work, believing that pure mathematics was an art form that should not be tainted by practical applications. Ironically, his contributions to analytic number theory now underpins modern cryptography.
It’s weirdly difficult to study something for years - no matter how abstract — and successfully avoid any practical applications!
This reeks of survivor bias and I'm interested to hear arguments why you're so confidence this is the case without resorting to the obviously very well-known survivors (as opposed to the thousands of research projects that ended apparently nowhere and long forgotten...)
For a modern version of the paper with some additional notes, see https://data.jigsaw.nl/Rado_1962_OnNonComputableFunctions_Re...
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.
[0] https://www.sligocki.com/2023/03/13/skelet-1-infinite.html
[1] https://discuss.bbchallenge.org/t/skelet-17-does-not-halt/18...
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.
Proving something in a theorem prover means the proof itself is an object constructed in the prover's language.
Nobody is suggesting this, and in this case, it was indeed "ported" to Coq from existing sketches.
The distinction here isn't between on-paper-first vs computer-first. The distinction here is between using a custom computer program to perform computations for a mostly-paper proof, versus taking an existing general-purpose theorem-checker (Coq, in this case) and writing down the entire proof in its language so it can check it.
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.
You mean the 2-state 4-color 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.
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.
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.
How do we know that there would be consistency issues or Σ₁-soundness issues?
Your claim about proof size categorizing n-state machine halting status is new to me. Do you have any links to read more about this?
The argument doesn't make sense to me. Rather it seems like more of a consequence of BB being incomputable in the first place. The proof sizes for each BB(n) aren't expected to be computable at all. There is necessarily a different theory for each n (or intervals of n where each theory applies with limits on each).
> So there isn't anything to support the position, other than the pure speculation that anything is physically achievable given enough time.
Something something burden of proof something. It would be extremely fascinating to have a conclusive argument that large BB numbers cannot be solved.
From Gödel's second incompleteness theorem, no consistent first-order recursively-axiomatizable theory (i.e., a theory that can have its proofs validated by a Turing machine) can prove its own consistency. Thus, to prove that your current theory (e.g., ZFC) is consistent, you must move to a stronger one (e.g., Aleph*). But then you can't prove the consistency of that without an even stronger theory, and so on. Thus, you end up with an infinite regression, and you can't ultimately prove the consistency of any of these theories.
> Your claim about proof size categorizing n-state machine halting status is new to me. Do you have any links to read more about this?
Not really, other than some of my own ramblings on the bbchallenge Discord server. But it's not that long:
Suppose that the longest-running n-state machine M can always be proven to halt using a proof of under f(n) symbols, where f(n) is some fixed computable function. Then, you could construct a TM that, given n, enumerates every valid proof of length less than f(n) symbols, and checks whether it shows that a particular n-state machine halts. The TM then simulates all of the proven halters to see how long they run. By assumption, the longest-running machine M is included in this list. So this TM can compute BB(n), which is an impossibility. Therefore, the function f(n) cannot exist.
As a corollary, since it's "uncomputably difficult" to prove that the longest-running machine halts at all, it's no less difficult to fully establish the value of BB(n).
There is similar issue with even ZFC and PA. It’s not really a dealbreaker imo.
> Suppose that the longest-running n-state machine M can always be proven to halt using a proof of under f(n) symbols, where f(n) is some fixed computable function. Then, you could construct a TM that, given n, enumerates every valid proof of length less than f(n) symbols
The issue with that argument is that the TM which enumerates every valid proof can’t exist in the first place.
If you fix an axiomatic theory, it’s already known that the theory has a limit.[1]
If every theory has a limit, you need countably infinitely many axiomatic theories together to prove BB(n) for all n. So there’s no TM which can even enumerate all the proofs, since a TM must have finite states, and thus can’t enumerate infinitely many proof systems.
(In fact for similar reasons I believe a Halt program, which has infinite states but which works for all TMs with finite states, platonically exists. It’s an emulator and an infinitely long busy beaver number lookup table. The diagonalization argument doesn’t apply, since the infinite Halt doesn’t accept itself as input.
This Halt would have countably many states since each busy beaver number is finitely calculated and there’s only countably many of them.)
So it’s not clear that f(n) is uncomputable. If f(n) is the symbol count and not the binary encoded length of the symbols, it even seems that it’s trivially bounded by some constant for all n. The proof could be one symbol the meaning of which is encoded in the theory.
It is a fascinating question though. I’m sure there is some function of axiomatic theory proof checker TM size and binary encoded proof length which does grow with n. It’s unclear if it would be uncomputable though.
The consequence of it being uncomputable is that the universe doesn’t have the resources to even encode the theory and/or represent it’s proofs.
In fact I suppose even as long as it grows at all, there would be a limit to BB(n) which can be possibly determined. Very fascinating
[1]: page 5 https://www.scottaaronson.com/papers/bb.pdf
We obtain PA from our basic intuitions about how the standard integers work, derived from empirical evidence. Everything past that involves increasing levels of intuition. So to continue it indefinitely, you must postulate an infinite amount of correct intuition, in some magical fashion that can never be captured in a computer. You can claim unlimited ingenuity all you want, but there's no a priori reason that it should indefinitely yield the truth, especially when it goes far, far past what our finite empiricism can provide.
We just haven't hit these limits yet, since very weak inductive theories are still sufficient for proving BB(5): we don't even need the full power of PA yet for our non-halting proofs. Thus why it looks like it should be so easy.
> If you fix an axiomatic theory, it’s already known that the theory has a limit.[1]
Not quite. Fix some consistent axiomatic theory T which proves PA. Then there will be infinitely many TMs which do not halt (in the standard model), but T cannot prove that they cannot halt, due to incompleteness. (Therefore T cannot settle the BB(n) question past a certain point, as Aaronson correctly says.)
But for every TM that does halt (in the standard model), T can prove that it halts, and the proof is to list out a trace of the TM's execution. Thus, every halting machine of every length has a halting proof in T.
The only benefit of a more powerful theory T is that it can "compress" this maximal BB(n)-sized proof into something more physically managable. But once we fix a certain T, we find (by my earlier argument) that it can only compress the proof so far, and the compressed size still must be an uncomputable function.
We can also see this by a forward argument, instead of by contradiction. Suppose that we'll accept any halting proof in a theory T. Then we can write a TM that lists through all proofs in T that are smaller than some bound N. (Notice that this is a finite set, since I've put an upper bound on it!) Then, for every proof that is a valid halting proof, the TM runs the corresponding machine. Then, the TM will halt, and its halting time will be greater than the halting time of any machine that can be proven to halt in T within N symbols. Set N to Graham's number (which is easily definable), and now the halting proof of the TM in T will not fit in our light cone.
(Notice how our TM clearly halts if T is Σ₁-sound! But since T cannot prove its own Σ₁-soundness, it doesn't have any way to prove our TM halting other than by the brute-force method.)
> In fact for similar reasons I believe a Halt program, which has infinite states but which works for all TMs with finite states, platonically exists. It’s an emulator and an infinitely long busy beaver number lookup table. The diagonalization argument doesn’t apply, since the infinite Halt doesn’t accept itself as input.
In that case, you just end up with the well-known oracle halting problem, where you equip a TM with access to this "infinite-state" machine. Then the problem is that you have a more powerful model of computation, but still with no way of solving its own halting problem. Much like how a consistent theory can only prove the consistency of weaker theories, not its own consistency.
> So it’s not clear that f(n) is uncomputable. If f(n) is the symbol count and not the binary encoded length of the symbols, it even seems that it’s trivially bounded by some constant for all n. The proof could be one symbol the meaning of which is encoded in the theory.
Of course I'm fixing a particular theory and a particular alphabet of constant size, the alternative would be absurd. The important part is about the ultimate behavior as n varies.
I’m just not convinced that fixing a theory and disallowing soundness axioms is any practical impairment for discovering busy beaver numbers.
Of course things get out of hand, but then why not collect evidence for soundness and let proofs avoid including an actual execution of the TM?
We don’t need a printout of grahams number computation to know a TM which computes it halts. why impose that here?
Has anybody done a compilation of such a PA proof generating/checking and simulating TM? It must have an enormous number of states and almost certainly wouldn’t be the BB of its cohort. Not including how it should be a similar thing where to prove it’s step count we shouldn’t need to emulate it.
Of course, the TM which "computes Graham's number" can trivially be proven to halt, without running it fully.
But it is much harder to show that the TM (let's call it M) which "computes Graham's number, then plugs it into a big halting-proof generator (in a fixed theory T) as an upper bound on the number of symbols, then executes each proven-halting TM in order" also halts.
The problem is, just because a theory T gives a halting proof for a machine, doesn't necessarily mean that it halts. (That is, T isn't necessarily Σ₁-sound.) And if that doesn't hold, then M might run into a "fake halter" that can be proven halting in T, but doesn't truly halt in the standard model. Thus, the only ways to show that M halts are to either establish that T is Σ₁-sound, which can only be done by appealing to a stronger theory, or to run through each of the Graham's number of proofs, which takes astronomically long.
This is similar to an instance of the Berry paradox: if you could easily prove that M halts, then it would have a relatively short halting proof within T. But then it would find its own halting proof, and simulate itself. But then it would simulate itself simulating itself, etc., etc., and never actually halt. So T wouldn't be Σ₁-sound, since M doesn't halt even though you proved that it halts. The only way out of this trap is if M doesn't simulate itself, i.e., it takes more than Graham's number of symbols in its own halting proof in T.
This appears to be an argument that almost every axiomatic theory isn't Σ₁-sound. So I can only ask a few things:
1. How do I even learn about Σ₁-soundness? I've tried for 30 minutes now to search for it.
2. Is my impression correct? Does this argument show that for ex ZFC is not Σ₁-sound? If not, which axiomatic theories exactly fall into this trap?
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.
It’s fascinating that the entire space of finite amounts of random gibberish contains every such stronger theory.
As a thought experiment it does well. Interestingly the Church-Turing thesis seems to exclude ingenuity. That is, it doesn’t try to say there aren’t functions on natural numbers which are uncomputable but can be calculated with ingenuity. Seems that a ton of people conflate those things.
ZFC and larger theories are only relevant for infinite objects, not finite objects like BB.
https://mathoverflow.net/a/26605
Furthermore, Scott aaronson jokingly conjectured that even BB(20) is independent of ZFC. That’s my reference here where we end up needing Aleph* for BB(18).
This is simply a test for if consciousness has infinite computational resources.
> 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.
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)?
For collatz, the empty input machine loops over all natural numbers and halts if it finds one which doesn’t eventually reach 1.
To prove that it never halts, you’d have to prove the collatz conjecture. Otherwise you’d have to find the smallest counter example of the collatz conjecture.
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...
> 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.
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?
If you can encode a problem in n states, and you know BB(n), then, as you say, you could use that to solve the problem. Trouble is, how do you know BB(n)? In reality, the only way to determine BB(n) is to go and solve all such problems; there isn't any other easier method that you can apply that would then let you get answers for these problems as a consequence.
So, the value of BB(n) will always be a summation of "we did all the hard work of solving all the n-state problems", not something you do separately to get those answers out.
Actually, just as you can do bijective base-b [0], you can also do bijective Zeckendorf (using 1 and 2 with no two consecutive 1s). Although, as happens with bijective binary, bijective Zeckendorf is closely tied to ordinary Zeckendorf, so it doesn't offer much new. But bijective dual Zeckendorf doesn't work -- lots of numbers can't be represented!
One more fun fact about Zeckendorf and dual Zeckendorf: Write n>0 in Zeckendorf, and count how many zeroes it ends in. This will be even if the dual Zeckendorf representation of n ends in a 1, and odd if it ends in a 0. Similarly, if you write n in dual Zeckendorf and count how many 1s it ends in, this will be even if the (ordinary) Zeckendorf representation ends in a 0 and odd if it ends in a 1.
This isn't disagreeing with what I was saying, though; I said incompleteness requires there is some such n, and then also said what we know about the record, I didn't say the record comes from a Gödel sentence.
Hope my comment addresses a common confusion despite its unnecessarily corrective tone.
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?
Insofar as we are given provability, we can solve halting.
Would you be brave enough to say the same thing about BB(6) or BB(7)?
Because that one needs to be solved for BB(6) already: https://wiki.bbchallenge.org/wiki/Antihydra
Considering how we made basically no real progress on it mathematically in a whole generation, solving BB(6) within the next decades would be a miracle, and I would bet a lot against it.
I can't see us EVER getting to BB(10), no matter how advanced humanity grows (and it would be meaninglessly large anyway).
I think 765 is just a huge overestimation based on the fact that it is somewhat straightforward to construct.
But yes, I'd be very surprised if Collatz were actually undecideable, even if it's well beyond the reach of current mathematics.
I agree with your statements about BB(10) and BB(6), but they just aren't very relevant. I agree those involve extremely difficult problems likely well beyond the reach of current mathematics, but I'd still be very surprised to find anything undecideable in there. There's a big difference between being truly undecideable and merely well beyond the reach of current mathematics!
(Also, the current record for undecideability is 745, or 765.)
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.
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
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.
- You can "flip" rules/initial tape values, trying to get a machine that can start with all 0s. This is not always possible. A trivial case where it is possible is a machine that wants to start with all 1s (just flip everything).
- You can insert more rules to set the tape up as needed. This makes it an BBn+x busy beaver candidate, but at least it's not your problem at BBn now.
The last way can get a bit complicated because you need to insert ad-hoc logic, since the tape is infinitely long. So it's not a just a case of inserting setup at the beginning of the program (because you can't setup an infinite tape in finite steps). Also to be fair this second point is a non-trivial assertion and requires proof that it is actually possible to do in every case. It is obviously possible to do for every machine that runs in finite steps though (since it can only consider a finite amount of tape).
Luckily actual computers have finite memory which is generally initialized with 0 anyways.
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.
> 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).
- 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.
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_%...
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.
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.
My feeling is that this trend cannot continue forever, and for infinitely many N they are different. If they are always the same, then you could find the steps champion just by finding the marks champion. This would be convenient, because as you pointed out, steps are more logically important, while marks are more practically important. But this feels too good to be true, and so it probably isn't.
For example (using run-length encoding), 1^n 0 1^m might become 1^(n-1) 0 1^(m+2)
When the rule is applied, the transformation is applied directly to the tape, generally by manipulating some count variables.
Now, how many machine steps does it take to apply this transformation? Well, TBH I'm not really sure. It seems kinda complicated, especially when the rules get more elaborate. If you are trying to run your simulator as fast as possible, you probably don't want to bother calculating it at all anyway, since you can always rerun the analysis at a more leisurely pace later.
So when I say that marks are "more practically important", I mean that marks are central to the operation of advanced simulators, whereas steps are a derived afterthought value.
Logically, the steps are more important, since they give you an easy method for solving the halting problem for the state/color class.
So far, the markiest programs are also the steppiest. My conjecture is that they will turn out to be different in infinitely many classes. If they were always the same, you would be able to get the logical primacy of steps just from working with marks. And that sounds too good to be true.
Under "Turing machines with 5 states and 2 symbols", there's no mention of the definitive results
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.
Personally, I am slightly leaning towards clean code being more powerful. Chaos is hard to control.
https://nickdrozd.github.io/2022/03/12/formal-theory-of-spag...
But yeah, it's an opinion. :)
This is not quite right. We're selecting for maximally long-running (or mark-producing, etc) programs. Whether those programs turn out to be "pathological" in some sense is an open question, and a question that can only be answered (to a limited extent) by observing and analyzing actual champion machines. Apriori, there's nothing we can say about what a champion program will do or what its code will be like. The Spaghetti Code Conjecture says that these champion programs will tend to be complicated. I have argued in the past that this may not be the case. It's entirely possible that the code of a champion program will be written more or less straightforwardly, and its long-running-ness comes from executing some bizarre and exotic mathematical function.
Actually, I think the more likely outcome is that some champions will be spaghetti and some will be clean. If they were all spaghetti or all clean, then that fact could be exploited to speed up the search process by discarding all programs not matching the property. And that sounds too good to be true, so it probably isn't. Maybe BB(8) champion is clean, BB(9) is spaghetti, etc, and there are no reliable trends.
Ah, now, that's thinking with pathologies.
my team has also proved this via our production codebase
> 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.