Talk: Busy beaver function

Archive 1

Poll

It seems that overall the term "busy beaver function" is more popular. Should we consider moving this page? -- vel! 19:23, September 5, 2015 (UTC)

<poll> Should we rename the article to "Busy beaver function"? Yes No </poll>

Weird alternative for S(n)

Define s(n,c) to be maximum shifts function for finite tape limited to c cells. If this finite-tape TM's head escapes cells, it halts. Is g(n) = s(n,n) uncomputable? And can s(n,c) from some c be uncomputable? Ikosarakt1 (talk ^ contribs) 05:18, September 19, 2015 (UTC)

There are only finitely many (exponentially many) possible tape configurations, so we know that the machine doesn't halt iff it repeats a configuration. We can check if this happens by simply keeping track of all configurations the machine reaches and looking for repetition. Hence we can decide whether these machines halt, which makes s(n,c) computable. Indeed, this function is upper bounded by number of possible configurations which, I believe, is \(2^c\cdot c\cdot n\) (each of c cells can take 2 values, there are c possible head positions and n possible states)

By the way, this is closely related to linear bounded automaton. LittlePeng9 (talk) 06:50, September 19, 2015 (UTC)

I was too slow and had a conflicting edit saying basically the same. A few things to add: your machine (I'll call it a Deterministic Bounded Automaton) can be thought of as a Turing machine where the input tape has the symbols [], symbolizing the left and right ends of the tape. For a TM to be an DBA, the TM's rules must restrict it from escaping or overwriting the end markers. Also, the input must be wrapped in these markers and [ has to be immediately to the left of the TM starting position.

LittlePeng9 gave an upper bound and a computability proof, but didn't comment on how easy it is to compute. The DBA halting problem is primitive recursive (since the simulation time is bounded), and so is \(s(n,c)\). -- vel! 07:20, September 19, 2015 (UTC)

Fun fact

BB(n) mod TREE(n) isn't computable function, but TREE(n) mod BB(n) is. Ikosarakt1 (talk ^ contribs) 04:01, September 23, 2015 (UTC)

Can you prove the first statement? For what I know, it is possible that for all n large enough BB(n) is divisible by TREE(n). LittlePeng9 (talk) 04:54, September 23, 2015 (UTC)

Which proof theory should I use? Ikosarakt1 (talk ^ contribs) 12:11, September 23, 2015 (UTC)

I suppose you are asking which axiomatic system to use. I would say that ZFC will be fine. LittlePeng9 (talk) 13:22, September 23, 2015 (UTC)

Sorry, I can make proof only in ZFC+"BB(n) mod TREE(n) is uncomputable". So yes, it's just a fun hypothesis. Ikosarakt1 (talk ^ contribs) 03:49, September 24, 2015 (UTC)

What if the theory you described is inconsistent? -- vel! 13:04, September 24, 2015 (UTC)

It would mean that BB(n) mod TREE(n) is computable. Ikosarakt1 (talk ^ contribs) 14:24, September 24, 2015 (UTC)

I think vell's point is that in this case your proof would be unsound, so that you would prove something that's not true.

(it's also possible that this theory is inconsistent because ZFC is, but that's beyond the point) LittlePeng9 (talk) 14:30, September 24, 2015 (UTC)

Shocking thoughts

I've heard about universal Turing machines which can simulate others. But that makes BB(n) even more crazy. Imagine, say, 100-state TM (A) which can simulate another, billion-state TM (B) which returns insane number of 1's (corresponding to its recursive level). Or even worse, TM (B) may simulate some TM (C) with even more states. In that case, I think even BB(100) is hopeless to be beaten by explicitly defined recursive function (not even saying about proving its value). 109.94.79.14 19:53, October 25, 2015 (UTC)

Universal turing machines (those that simulate others) don't work exactly like that. They never halt, and are built such that they eventually simulate all possible turing machines. If a turing machine could simulate a turing machine with a greater number of states (and an equal/greater amount of colors) than the function would clearly be ill defined. For example, we'd have BB(100) equal at least BB(10^6), which would be equal to at least BB(10^12), which would be equal to at least BB(10^25), which would be equal to... and so forth till literally infinity. Direct simulation of turing machines with smaller amount of states is possible though (and with 2 tapes it's not even very hard), and you can see an example of that on my blog post here. Maybe called Googology Noob (talk) 11:07, January 22, 2016 (UTC)

Actually, a Turing machine (A) with \(n\) states can simulate a machine (B) with \(m>n\) states. But here is the catch: (A) must operate with a specific input, not just empty input. An example is this machine of mine, which in principle can simulate a binary Turing machine with arbitrary number of states (the machine itself isn't binary, but there are standard ways of turning it into a binary TM). This doesn't make BB ill-defined, because we need to encode whole machine (B) into a string which would be a prefix to the input. LittlePeng9 (talk) 12:40, January 22, 2016 (UTC)

In theory, you can have turing machines that can simulate any other turing machine given the right input. The problem is, with the small ones, the input is infinitely wide, and even with larger ones, having the machine that would write the input for the universal machine would usually take at least as many states as the original turing machine does. Tomtom2357 (talk) 08:55, January 25, 2016 (UTC)

LittlePeng9: That sounds interesting, but your link directs to the binary palindrome checker. Maybe called Googology Noob (talk) 15:08, January 30, 2016 (UTC)

New Bounds

I believe I have shown some new bounds of the Busy Beaver function using an accelerated FGH (f_0(n)=n+1, f_a+1(n)=f_aⁿ⁺²(n+2), f_a(n)=f_a[n+2](n+1)) here, mainly that \(\Sigma\)(25)>f_w2(3) and that \(\Sigma\)(27)>~f_w^2(12) (though obviously more bounds can be made on that). Can I put those bounds here? Maybe called Googology Noob (talk) 07:16, January 9, 2016 (UTC)

Your bound isn't a bound for \(\Sigma(27)\), but rather \(\Sigma(27,6)\), which is a multi-color variant of BB function (the symbols used are

_,1,w,W,i,|,l

, did I miss anything by a chance?). Although this is some bound, I honestly don't think it's of enough relevance to put it on this page - we tend to put multi-color bounds only if they beat some "milestone" in FGH. LittlePeng9 (talk) 08:05, January 9, 2016 (UTC)

It uses all of those symbols, but not every symbol is used by each state (Obviously such a bound for \(\Sigma(27,6)\) would not be much). Each state uses exactly 2 colors. Maybe called Googology Noob (talk) 12:39, January 9, 2016 (UTC)

That doesn't matter - for \(\Sigma (m,n)\) you can use no more than \(n\) colors in total. Deedlit11 (talk) 23:20, January 9, 2016 (UTC)

Oh, OK. Somehow I didn't realize that. Maybe called Googology Noob (talk) 06:07, January 10, 2016 (UTC)

Other numbers estimated using BB(n)?

I think for computable numbers the Busy Beaver function would be a good way to possibly envision their size. I know that BB(18) is greater than Graham's Number, and I heard that Loader's Number is somewhere around BB(160), but what abuot the following?

- Zootzootplex

- TREE(3)

- SGC(13)

- Tetrathoth

- Gongulus

- The various Fish numbers (aside from 4 and 7)

- Kungulus

- Meameamealokkapoowa oompa

Does anyone have any estimates? ^{—Preceding unsigned comment added by 148.75.113.243 (talk • contribs) 10:12, September 6, 2016 (UTC)}

There's no way it was proven that Loader's number is somewhere around BB(160). What they meant was probably that some number around 160 is the smallest number such that some people figured out a proof that it's larger than Loader's number. It can be proven from the Church Turing thesis that if for each positive integer n, you will eventually state a lower bound for BB(n) and only do it once, then for some t, BB(n) is larger than the lower bound for it that you will state for all n > t. The same goes for the frantic frog function S(n). There probably exists a 6 state 2 color turing machine that can some people have the resources to prove doesn't halt after TREE(3) steps but hasn't been proven to eventually halt and it can probably be proven that there's no proof in some very high level proof system that it never halts. If it does eventually halt, then S(6) is a whole lot larger than it's current proven lower bound. Although we probably don't have the resources to think of a mathematical proof of it, it can probably be proven from the Church Turing thesis that that turing machine doesn't even halt after a number of steps equal to Loader's number. Blackbombchu (talk) 18:40, September 6, 2016 (UTC)

1. Zootzootplex<googolplex^^googolplex. I *guess* it would be beaten by BB(8), or at least by BB(9). 

It has been proven to be beaten by BB(12) by Milton Green.

2. TREE(3). This is a bit of a hard one, since it is even hard to say where TREE(3) falls. If you look to the best proven bounds, it is going to be in the thousands, if someone actually proved it formally.

3. SCG(13). Also a hard one, for the same reasons as for TREE(3). I'm not going to guess on those, since it would most likely be wrong. It could be as low as, say, BB(12), or as high as BB(50). Of course, this is a guess - I can't prove it can't be BB(11) or BB(51).

4. Tetrathoth is beaten by Hydra(1907), and hence BB(85) would be larger than it.

5. Gongulus is also beaten by BB(85), I don't know any better bound.

6. Fish numbers -  Fish numbers 1,2,3 are beaten by BB(85). I think with some work I could show BB(100) beats Fish number 5. For Fish number 6, the best known is probably in the thousands, based on color reduction methods.

7. Kungulus, Meameamealokkapoowa oompa: I won't comment on these since it largely depends on the definitions chosen beyond tetrational arrays. Wythagoras (talk) 19:46, September 6, 2016 (UTC)

R function is uncomputable. AarexWikia04 - 00:00, September 7, 2016 (UTC)

How is the R function uncomputable? Are you talking about something other than the function defined by Hyp cos? If so, how is this have anything to do with the above comment? -- ☁ I want more clouds! ⛅ 13:43, September 7, 2016 (UTC)

Oh wait, Fish Number 4 and Fish Number 7 are uncomputable. Also, we need better bound for S(8). AarexWikia04 - 14:56, September 7, 2016 (UTC)

How can TREE(3) be in the thousands and SGC(13) be between 12 and 50, if SGC(13) is bigger than TREE(3)?

Also, BB(1000) is listed on the main website as being higher than any of the numbers I asked about.

For the Bowers numbers, what would be their approximate BB values if you use the most generous definition (the one that gives the highest numbers)? ^{—Preceding unsigned comment added by 148.75.113.243 (talk • contribs) 12:34, September 7, 2016‎ (UTC)}

About TREE(3) and SCG(13): The thousand part is talking about bounds that are actually feasible to prove, while "12 to 50" means what we guess where the actual value falls in.

About kungulus: in one interpretation (which seems to be advocated by Bowers himself), kungulus' growth rate is aroung \(\Gamma_0\) (the Feferman–Schutte ordinal), which is smaller than the Bachmann-Howard ordinal. There is a bound for Bachmann-Howard ordinal using 81 states and 10 colors, which is around a few hundred states and 2 colors using the color reduction technique. I still won't comment on meameamealokkapoowa oompa though, since one definition puts it in an incredibly high growth rate that I can't understand it. -- ☁ I want more clouds! ⛅ 13:53, September 7, 2016 (UTC)

Rayo's number

How many states do we need to beat Rayo's number? Googol, Googolplex or Loader's number? --84.61.148.21 17:52, September 7, 2016 (UTC)

Since Rayo's number is bigger than Sigma(), there is no way to result it. AarexWikia04 - 19:55, September 7, 2016 (UTC)

Rayo's number: This needs extremely many states, something that is not really much smaller than Rayo's number.

Googol: 6 is enough, 5 is conjectured to be not enough.

Googolplex: 7 is enough and I would personally conjecture 6 to be enough as well.

Loader's number: It is not known, but probably 1,000 states is sufficient.

Wythagoras (talk) 14:27, September 9, 2016 (UTC)

\(\Sigma\) might be faster than we suspect

I call Turing machine (TM) "semi-universal" if it can simulate some other TM (particularly with larger number of states) with some input.

Now imagine if there is 10-state TM which simulates some 1000-state TM which computes some fast-growing recursive function with sufficiently large input.

What if \(\Sigma(10)\) is already larger that anything what we defined so far using computable function?

Moreover, there might be "chains" of simulations: that 1000-state TM might be itself a simulator, for, say, \(10^{100}\)-state TM! (o_O) 185.137.18.186 05:13, November 14, 2016 (UTC)

Indeed. Just because we could build TM of some power with a bunch of states, it doesn't mean we can claim it isnt beat by a much lower numbered BB champion. Chronolegends (talk) 05:25, November 14, 2016 (UTC)

Although this is possible, it is rather unlikely that this approach will lead us to a new lower bounds on BB numbers. The reason for why I think so is that, in order for this to be feasible, the additional input allowing the 10-state TM to simulate the 1000-state TM would have to be relatively short, which would mean that, whatever the 1000-state TM was to do, it was terribly wasteful at it, given that a machine with so many states fewer was able to do the same job. In practice, if this were done, one would probably that the 1000-state TM was a less efficient version of the 10-state TM, rather than that the 10-state TM is a more efficient version of 1000-state TM. So all in all, I just think "semi-universal" TMs are plain unfeasible.

As for the remark about \(\Sigma(10)\) - heck, we don't even know this about \(\Sigma(5)\)! As unlikely as it seems, it might be the case that one of the unresolved 5-state TMs produces output vastly larger than any known computable notation. LittlePeng9 (talk) 09:29, November 14, 2016 (UTC)

Let's say, this 10^100-state TM simulates a Meameamealokkapoowa oompa-state TM. 80.98.179.160 11:22, January 12, 2018 (UTC)

It is by all means possible - there is a meameamealokkapoowa oompa-state TM which prints ones forever, and there are 10^100-state TMs which simulate that behavior. LittlePeng9 (talk) 15:11, January 12, 2018 (UTC)

By the way, there is an idea about how to improve our current bounds for \(\Sigma(n)\). One can pick some UTM, like this. Then waste some states to write input correspondent to some known powerful TM, like this. 185.137.18.186 07:58, November 14, 2016 (UTC)

First, let me point out that using Wolfram's (2,3) TM would be probably the worst idea to simulate another TM one can think of. There are a few reasons for that, among them the fact that this TM needs an infinite, nonrepeating input and the fact that it never halts. To be honest, I don't even know under what measure this machine is universal (and it seems community doesn't know either...).

There are some proper UTMs which could be actually used for this purpose there are small 2-symbol UTMs (there is one with 19 states, there is also one with 18 states, but it lacks a halting instruction too). However, these UTMs require input of considerable size, which would take a long time to produce (we'd have to waste a lot of states), causing us to spend more states than the original machine had (using Baiocchi's machine and the hydra machine you've linked we'd probably only get a lower bound for \(\Sigma(n)\) for \(n\) on the order of thousands if not tens of thousands.

On a positive side though, such UTMs can be used for less specific bounds, for example to prove that \(\Sigma(n)\) outgrows any computable function. There are easier ways to do this as well, but still :) LittlePeng9 (talk) 09:29, November 14, 2016 (UTC)

Wait what, this must be fastest! AarexWikia04 - 12:33, November 14, 2016 (UTC)

What must be fastest? LittlePeng9 (talk) 13:04, November 14, 2016 (UTC)

Basic BB question

Why does a 2 state busy beaver produce more 1s than a 1 state busy beaver? What is the difference between a 1 state and 2 state busy beaver? —Preceding unsigned comment added by 172.58.111.42 (talk • contribs)

Do you know the definition of a Turing machine? Do you know what's the difference between 1-state and 2-state Turing machine? LittlePeng9 (talk) 08:13, November 18, 2016 (UTC)

BB(n) and ordinal

Does anyone know which BB(n) has the large veblen ordinal? —Preceding unsigned comment added by 172.58.104.127 (talk • contribs)

What does it mean that a number has an ordinal? LittlePeng9 (talk) 09:34, November 19, 2016 (UTC)

So the Busy Beaver function doesn't use recursion levels? So the turing machine states are not ordinals or recursion levels? —Preceding unsigned comment added by 172.58.104.162 (talk • contribs)

Correct. LittlePeng9 (talk) 18:15, November 19, 2016 (UTC)

Golapulus

Does anyone know which BB(n) is larger than Golapulus? —Preceding unsigned comment added by 172.58.107.103 (talk • contribs)

You mean one of these Bowers' numbers which aren't well-defined? Your question is hard to answer, because golapulus is, well, not well-defined. LittlePeng9 (talk) 19:45, November 20, 2016 (UTC)

It is well-defined!! 80.98.179.160 11:24, January 12, 2018 (UTC)

No it's not!!! Rpakr (talk) 15:09, January 12, 2018 (UTC)

Okay. —Preceding unsigned comment added by 172.58.107.103 (talk • contribs)

I suppose that n = 100 is enough, but it might be even smaller. 109.236.81.168 20:38, November 20, 2016 (UTC)

How does the Xi function go beyond the Busy Beaver function? —Preceding unsigned comment added by 172.58.110.148 (talk • contribs)

Did you read the article on Xi function before asking that? it has the answer you are looking for in part 1.2 "SKIΩ calculus" Chronolegends (talk) 08:11, November 22, 2016 (UTC)

SKI calculus is essentially equivalent in strength to Turing machines, so we expect the xi function without Ω combinator to be, in a way, comparable to BB function. But then adding Ω combinator makes SKIΩ calculus stronger than TMs, so xi function is stronger than BB. LittlePeng9 (talk) 08:27, November 22, 2016 (UTC)

Two questions

Does anyone know how many possible 100-state Turing machines there are? How is it a proven fact that BB(18) is larger than Graham's number? —Preceding unsigned comment added by 208.54.83.217 (talk • contribs)

First question: please make sure you have read the whole article before asking such a question. Second question: a machine has been constructed which achieves an output larger than Graham's number with 18 states. LittlePeng9 (talk) 20:46, November 23, 2016 (UTC)

The game

What is the difference between the Busy Beaver function and the Busy Beaver Game? —Preceding unsigned comment added by 172.56.15.26 (talk • contribs)

The busy beaver game is the problem of finding the values of the busy beaver function. LittlePeng9 (talk) 17:10, November 25, 2016 (UTC)

Okay. Thanks. —Preceding unsigned comment added by 172.56.14.229 (talk • contribs)

Why is it so strong?

What is the main reason the busy beaver function is uncomputable? Is it because busy beavers are turing machines themselves? —Preceding unsigned comment added by 208.54.83.250 (talk • contribs)

uncomputable just means there is no algorithm  that can compute it

Chronolegends (talk) 01:27, November 26, 2016 (UTC)

The main reason is the following idea: whatever computable function \(f(n)\) you choose (assume its increasing for simplicity), there is a Turing machine M which computes it. By adding a few states we can create a machine M' (with, say, \(C\) states) which computes \(f(2n)\). From there you construct, for each \(k\), a machine M'_k with k more states which computes \(f(2k)\) from the empty input, so \(BB(C+k)\geq f(2k)\). For large \(k\) this gives \(BB(C+k)\geq f(2k)>f(C+k)\).

Recap of the idea: for any computable function, we can create amachine computing a faster growing function and use it to give lower bounds for busy beaver. LittlePeng9 (talk) 09:21, November 26, 2016 (UTC)

Okay. Cool.

How does the Busy Beaver function relate to first-order logic? What are axioms for the Busy Beaver function? —Preceding unsigned comment added by 208.54.83.161 (talk • contribs)

There is no such thing as axioms for the BB function as far as I know. There is also no direct relation between BB function and first-order logic, but BB is expressible in some first order logics, like first-order arithmetic or FOST. LittlePeng9 (talk) 18:52, November 26, 2016 (UTC)

Oracle TMs

Does a n-state oracle turing machine output more 1s than a n-state regular turing machine? —Preceding unsigned comment added by 172.58.107.79 (talk • contribs)

Depends on an oracle and the TM - there are n-state oracle TMs which output no 1s, which is less than what some regular TMs output. But I suppose the question is more about oracle busy beaver vs regular busy beaver, in which case the answer is yes, at least for most oracles. The precise answer probably depends on the oracle and on how precisely oracle calls are implemented, but with, for example, a halting problem oracle, the corresponding oracle BB definitely outgrows the regular BB. LittlePeng9 (talk) 18:05, November 28, 2016 (UTC)

Section break

What is the difference between BB(n), BB(k), and BB(2k)? —Preceding unsigned comment added by 172.58.104.199 (talk • contribs)

What is the difference between n², k², and (2k)²? -- vel! 02:21, December 2, 2016 (UTC)

Is a N-state the same as a program? What is the difference between a N-state and a function? I know the Busy Beaver function is uncomputable, but if you were to try to compute the Busy Beaver function; how would the input look like? —Preceding unsigned comment added by 172.56.7.253 (talk • contribs)

"N-state" is a property of a Turing machine; every Turing machine uses a certain number of states, and the more that you can use, the more complicated your machine can be and the larger the class of functions you can compute.

It looks like you need to learn how a Turing machine works; the Wikipedia page is a good start. There are many examples of Turing machines at [1]. To compute BB(n), the idea is to try to program a machine that outputs as many 1's as possible when you start with a tape consisting of all 0's, and you have n states available. Deedlit11 (talk) 22:28, December 2, 2016 (UTC)

Does an oracle Turing machine go beyond BB(BB(BB(...n times...BB(BB(BB(n)))? In other words does a N-state oracle Turing machine output more 1s than the Busy Beaver function (N-state regular Turing machines) outputs being iterated into itself multiple times? —Preceding unsigned comment added by 172.56.7.253 (talk • contribs)

It depends on the oracle being used. The obvious next step after regular TMs is to add an oracle for either the Halting function or the Busy Beaver. (It doesn't much matter which since you can compute one from the other.) We can call these machines TM₂ machines, and then define BB₂(n) to be the largest number of ones outputted by a TM₂ machine with at most n states, starting from a tape of all zeroes.

Just as the BB(n) function will beat out any function computable by a normal algorithm, the BB₂(n) function will beat out any function computable by a algorithm where you are allowed to call the BB(n) function. Since "apply the BB function n times starting from n" is a valid algorithm, BB₂(n) will eventually dominate BBⁿ(n). You can of course go further, creating a long ordinal hierarchy starting from BB(n), and the result will still be computable from BB(n), and therefore beaten by BB₂(n).

By the way, please sign your posts, so the rest of us don't have to. Deedlit11 (talk) 09:17, December 3, 2016 (UTC)

How many more 1s would a 3-state oracle Turing machine output than a 3-state regular Turing machine? How do i sign my posts? I don't know how to do that. —Preceding unsigned comment added by 172.56.15.62 (talk • contribs)

It depends on the precise implementation of an oracle machine (and of course the oracle being used). Unfortunately, while there is a more or less standard implementation of the normal Turing machine (not universal, but the first few Busy Beaver numbers have been calculated for it so it's what we go with), there is no standard implementation for a machine with an oracle for the Halting function or Busy Beaver function. I can say a couple of things though: I imagine you will get higher numbers for oracle machines with the Busy Beaver function than with the Halting function, simply because it is an immediate fast-growing function. Another thing to worry about is how strictly defined the input parameters are for the oracle call. For example, if you require that to input n into the oracle, you must have a contiguous string of n 1's with the head at the leftmost 1, than with say two free states I don't know if you could input a number greater than n=2. On the other hand, if you don't require that the head be at the leftmost position of the string of 1's, then one can create a string of four 1's with two free states, so we could then make an oracle call and return BB(4) = 13. So it all depends.

To sign your posts, simply put four ~'s at the end of your post. Deedlit11 (talk) 00:43, December 6, 2016 (UTC)

What are inputs for the Busy Beaver function? How does an oracle Turing machine output more 1s by solving the halting problem? What is the difference between Turing machines that are Busy beavers and Turing machines that are not Busy beavers?172.58.96.120 00:14, December 15, 2016 (UTC)

Turing machines come with a tape, so whatever is written on the tape when the Turing machine starts running is the "input" for the Turing machine. When there is more than one tape, usually one will be labelled the "input tape".

If you have an oracle for the halting problem, then you can compute the Busy Beaver function. Given an input n, just search through all Turing machines with n states and check whether they halt or not. For the TMs that halt, run them through to completion, and count the number of steps or symbols printed. Keep track of the maximum score. After you have run through all possible Turing machines, you will have the Busy Beaver number.

Since you can compute BB(n), you can also compute BB(BB(n)), or BB^n(n), or BB^(BB(n))(n), or much faster growing functions starting from BB(n). So the BB function for Turing machines with an oracle for the halting problem grows very much faster than the normal BB unction.

A Busy Beaver Turing machine is simply a Turing machine that prints the most 1's. Deedlit11 (talk) 02:18, December 15, 2016 (UTC)

Okay. Thanks for the information. If Turing machine levels w+w are possible (nth-order turing machines, like oracle turing machines are 2nd-order turing machines) then are TMLVO (Large Veblen Ordinal-order turing machines) possible? Like TMw+w.172.58.97.143 03:25, December 15, 2016 (UTC)

Yes. You can have an \(\alpha\)th level Turing machine for any \(\alpha\) that you construct an ordinal notation for, so you can go beyond \(\omega_1^{\text{CK}}\). Deedlit11 (talk) 08:57, December 15, 2016 (UTC)

Are Turing machine states inserted or attached to Turing machines or are they built into Turing machines? Let's say for example, a 3-state Turing machine. Is the 3-state inserted into the Turing machine, so that the Turing machine can operate or is the 3-state already built into the Turing machine?172.58.110.201 21:50, December 15, 2016 (UTC)

A Turing machine is an abstract computing machine, and states are part of it. Really, you should read the Wikipedia entry Turing machine, and possibly google other intros to get an understanding. Deedlit11 (talk) 01:14, December 16, 2016 (UTC)

Could Turing machines compute Hypcos's R function and Harvey Friedman's TREE sequence?172.58.107.111 12:58, December 19, 2016 (UTC)

Those are both computable functions, so yes. Deedlit11 (talk) 01:05, December 20, 2016 (UTC)

How do you simulate a Turing machine?172.58.96.26 15:04, December 22, 2016 (UTC)

I recommend this simulator. LittlePeng9 (talk) 16:00, December 22, 2016 (UTC)

Which is more powerful? Infinite time Turing machines (ITTMs) or Rayo's function? What is the difference between regular Turing machines and Infinite time Turing machines?172.58.104.62 22:05, December 22, 2016 (UTC)

Rayo's function is stronger than any function computable by ITTMs.

For ITTMs, we allow the machines to run for an infinite period of time, and then we essentially let them continue the computation, and again and again. LittlePeng9 (talk) 22:22, December 22, 2016 (UTC)

So does an Infinite time Turing machine use an infinite amount of states, the same way a regular Turing machine uses an infinite amount of memory tape?172.58.105.180 00:46, December 23, 2016 (UTC)

No, that would be too powerful. With an infinite number of states, you can easily make any output tape you wanted. Deedlit11 (talk) 02:28, December 23, 2016 (UTC)

Which number is larger? BB_Rayo's numberth order(Rayo's number) or Rayo(Rayo(10^100))?208.54.83.214 16:53, December 24, 2016 (UTC)

To define BB_{Rayo(10^100)}(Rayo(10^100)), you can define Rayo(10^100), which takes 10^100 symbols in FOST, and then define BB_n(m), which I am not about to do but it won't take that many symbols (certainly less than a million, I would think). So BB_{Rayo(10^100)}(Rayo(10^100)) will be less than Rayo(10^100 + million), so it will certainly be lessthan Rayo(Rayo(10^100)). Deedlit11 (talk) 12:28, December 25, 2016 (UTC)

Okay. Thanks for the information. Merry Christmas!172.56.15.35 16:27, December 25, 2016 (UTC)

Is BB(Gongulus) possible? What is the difference between Infinite time Turing machines (ITTMs) and Universal Turing machines? Which is more powerful?172.58.110.255 23:09, December 28, 2016 (UTC)

BB(Gongulus) is certainly a well-defined number. (Assuming that you believe in the Busy Beaver function, of course.) Universal Turing machines are no more powerful or broad a concept that regular Turing machines. A universal Turing machine is a single Turing machine that can emulate any Turing machine by modification of the input tape. So both a "universal Turing machine" and "Turing machines" in general lead to the same class of computable functions, it's just that for the first we "program" the function via the input tape, whereas for the second we program the function by modifying the Turing machine itself.

Infinite time Turing machines are a far more powerful construction than Turing machines, leading to a much larger class of computable functions. Deedlit11 (talk) 06:41, January 12, 2017 (UTC)

If a Turing machine were to compute a{{1}} b how would the input and output look like?172.58.96.108 01:37, January 16, 2017 (UTC)

Does BB(n)>f(n) (computable/recursive functions) in the same way multiplication>addition? In other words does the Busy Beaver function eventually dominates all computable/recursive functions in the same way multiplication eventually dominates all addition?172.58.105.59 21:47, January 19, 2017 (UTC)

I'm not sure what you mean by your first question, what is a{{1}}b?

For your second question, a function f(n) eventually dominates a function g(n) if there exists a natural number N such that for all n > N, f(n) > g(n). Put another way f(n) is bigger than g(n) for all but finitely many values of n. So for any computable function g(n), BB(n) is bigger than g(n) for all but finitely many n. Deedlit11 (talk) 06:57, February 3, 2017 (UTC)

What is the difference between stack frames and states?172.56.14.33 21:50, February 15, 2017 (UTC)

New bounds for \(\Sigma(n)\)?

The busy beaver hunter Shawn Ligocki wrote an article on bounds for \(\Sigma(n)\). So, is it really has been proved that \(\Sigma(9) > 10 \uparrow\uparrow 28\) and \(\Sigma(11) > 3 \uparrow\uparrow\uparrow 720618962331271\)? Ikosarakt1 (talk ^ contribs) 14:56, August 28, 2013 (UTC)

Well, a better question, are Milton Greens bounds correct? I can't let the TM of \(\Sigma(6)\) work. Can someone check? Wythagoras (talk) 15:30, August 28, 2013 (UTC)

This is a late answer, but yes, they appear to be correct. I ran Green's 6-state (resp. 7-state) G-machine that outputs his BB_6 (resp. BB_7) , and found that it works just as claimed, halting in 436 steps (resp. 197700005 steps), leaving 35 ones (rep. 22961 ones) on the tape. These numbers agree exactly with those computed from his recurrence relations. Res0001 (talk) 21:08, May 21, 2017 (UTC)

Turing machines and computability

What is the main reason Turing machines can compute any Computable function? Is it because of the infinite memory tape Turing machines have? Is it because of infinite memory tape; the main reason why the Busy Beaver function grows faster than any Computable function? Is infinite memory tape the main reason why the Busy Beaver function eventually dominates all Computable functions?172.58.105.141 11:54, June 20, 2017 (UTC)

"What is the main reason Turing machines can compute any Computable function?" The definition of a computable function.

The underlying reason for BB dominating all computable functions is the fact that for any Turing machine computing some function, there is a Turing machine which computes a faster growing function. LittlePeng9 (talk) 12:18, June 20, 2017 (UTC)

Is a Turing machine the same as an algorithm?172.58.67.92 21:01, June 23, 2017 (UTC)

I read on the Turing machine wikipedia page that a Turing machine is an Abstract machine. Does this mean that a Turing machine can be either an actual electronic model that a person or persons build or something you write down on a piece of paper? In other words, can a Turing machine simply be represented by drawing it on a piece of paper or do you have to build an actual electronic model in order for a Turing machine to be a Turing machine?2607:FB90:5C25:9FEB:BBD7:8217:BE1B:11D5 21:52, June 28, 2017 (UTC)

Why do the values of the Busy Beaver function get so large so quick? What is the main reason?2607:FB90:945F:C94B:5885:1CDE:9E61:D3EC 17:05, July 18, 2017 (UTC)

According to a Googology user who I forgot the name of, the busy beaver inverse of Graham's number is 18 ( maybe even 15, because it said \Sigma(18) > G ). I also found that TREE(3)'s busy beaver inverse is writable in decimal notation, but is been said to be in the hundreds or thousands. ^{—Preceding unsigned comment added by 222.97.145.127 (talk • contribs) 21:56, October 13, 2017 (UTC)}

How does the Church-Kleene ordinal relate to Turing machines? How does the Halting problem relate to the growth rate of the Busy beaver function?2607:FB90:44A7:AB83:3158:AE14:F596:EE1B 18:36, November 19, 2017 (UTC)

\(\Sigma(1919)\)

"There is a 1919-state 2-color TM, and whether it halts is independent of ZFC", then how do we get "\(\Sigma(1919)\) is independent of ZFC"? {hyp/^,cos} (talk) 13:55, December 9, 2017 (UTC)

From just the former statement I don't think the latter follows directly. However, I strongly believe that the machine constructed for the former statement, when halting, will leave a large amount of nonblank characters (unless it does clean the tape, I haven't looked through the code). If ZFC could prove an upper bound for the value of \(\Sigma(1919)\), then we would know an upper bound for the shortest inconsistency proof of ZFC, if there is one, and Godel's theorems guarantee it's not possible.

What we do get directly is that the value of \(S(1919)\) cannot be established in ZFC. LittlePeng9 (talk) 15:35, December 9, 2017 (UTC)

Second Order Busy Beaver Function

I have seen a file on this site with bounds for the second order busy beaver function. It is here: https://googology.wikia.com/wiki/User:Wythagoras/Rado%27s_sigma_function/Oracle_TMs?useskin=oasis I would like to be able to run this program on my computer and post the results on my blog post so everyone else can see it. I am also calling for research to be done on the second order busy beaver function. WaxPlanck (talk) 21:09, October 6, 2018 (UTC)

Formal definition as a set

Is it possible to define \(\Sigma(n)\) as the largest number in the set \(T = \{n_1,n_2,\cdots,n_k\}\) of all outputs of some \(n\)-state Turing machine? Note that the number of elements in \(T\) grows only "a bit" faster than exponentially. Triakula (talk) 08:28, January 13, 2020 (UTC)

Do you mean that you want to know whether the set \(T\) is well-defined given an input \(n\)? Then the answer is yes.

p-adic 10:24, January 13, 2020 (UTC)

I meant that there might be some other definition of \(\Sigma(n)\) which also makes clarity about that the number of numbers definable by \(n\)-state Turing machine is not so big and can be bounded below hyper-exponential function respect to \(n\). If we would physically have an oracle, then it wouldn't be too long to compute \(\Sigma(n)\). Triakula (talk) 10:54, January 13, 2020 (UTC)

Is there any difference between the original definition and your definition? (I guess that you just omitted other conditions such as 2-colouredness.) If you are considering the set of all halting n-state Turing machines versus the set of all outputs of them?, then those two are not so different because there are essentially at most (4n)^{2n} n-state Turing machines.

p-adic 12:50, January 13, 2020 (UTC)

The difference is in the definition of set: existence of this set \(T\) and its modest size is omitted in the article. But it may be important to show that despite of the big size of \(\Sigma(n)\), the pool of numbers that can be defined using \(n\)-state TM is very small. Triakula (talk) 13:10, January 13, 2020 (UTC)

I see. Thank you.

p-adic 13:23, January 13, 2020 (UTC)

Σ(2), Σ(3) and Σ(4)

Statements that Σ(2) = 4, Σ(3) = 6 and Σ(4) = 13 are believed to be proven but I can't find papers where it has been proven. I appreciate if anyone would point where are the proofs precisely. Triakula (talk) 14:08, January 17, 2020 (UTC)

ZFC

Can \(\Sigma(n)\) be defined in ZFC? If it can be defined in general, then how \(\Sigma(10 \uparrow\uparrow 10)\) is supposed to escape ZFC? Triakula (talk) 08:57, February 3, 2020 (UTC)

It is definable. Please see also this section. I do not understand the meaning of "escape" in the second question. Could you explain more about the question?

p-adic 09:23, February 3, 2020 (UTC)

I mean, why the formula \(\exists m : m = \Sigma(10 \uparrow\uparrow 10)\) is not within ZFC while \(\Sigma(n)\) can be totally defined over all natural numbers in ZFC? Triakula (talk) 09:33, February 3, 2020 (UTC)

It is a valid formula in ZFC set theory. I guess that you comfound the validity of the formula with the provability of a formula of the form m = Σ(10↑↑10). If m is a meta natural number in the sense here, then I guess that it is unprovable, because the value of Σ(10↑↑10) is perhaps undecidable. (The provability of such an equality means the decidability of the value.)

p-adic 09:47, February 3, 2020 (UTC)

Does it mean that the formal proof of \(\exists m : m = \Sigma(10 \uparrow\uparrow 10)\) would be just unreasonably long yet finite, or we can't express it with a finite number of symbols in ZFC? Triakula (talk) 09:54, February 3, 2020 (UTC)

No. The equantified formula which you wrote is provable quite shortly, because the instance of m can be substituted by Σ(10↑↑10) itself. The point is that I am referring to the closed formulae 0 = Σ(10↑↑10), 1 = Σ(10↑↑10), 2 = Σ(10↑↑10), …, 1000000000 = Σ(10↑↑10), and so on. It is not a single formula given by quantifying m, but is an infinite set of formulae indexed by meta natural numbers.

p-adic 10:01, February 3, 2020 (UTC)

Okay, thanks. Triakula (talk) 10:05, February 3, 2020 (UTC)

For example, let \(n\) denote \(\textrm{min}\{m\in\omega:|\mathcal P\aleph_0|\ge\aleph_m\}\). This is definable in ZFC, but ZFC can't prove \(\ulcorner\overline n=m\urcorner\) for any meta-natural-number \(n\) (where we define \(\overline 0:=0\) and \(\overline{n+1}=(\overline n)\cup\{\overline n\}\), I hope I've handled meta-natural-numbers properly) C7X (talk) 00:35, 14 February 2021 (UTC)

Uncomputability vs undecidability

Although the uncomputability of a number has never been formalised, I think that it is reasonable to express the undecidability of the equality with standard natural numbers as the uncomputability, because one formulation of the computability is compatible with it. How about writing both the uncomputability and the undecidability? (It is not so ambiguous then, as it clarifies "in the sense blah-blah".)

p-adic 16:31, 25 January 2021 (UTC)

I was under the impression that even the expression "\(\Sigma(3)\)" is uncomputable, because the uncomputability of an expression relies on the uncomputability of functions in its definition C7X (talk) 17:01, 25 January 2021 (UTC)

You are right. The notion of uncomputability of a number is ambiguous and heavily depends on googologists, Therefore the article clarified "in the sense blah-blah". (I note that the constant map f(x) = Σ(3) is computable and can be abbreviated to Σ(3). In this way, there is currently no almighty formulation of the uncomputabilty of a number.)

p-adic 22:51, 25 January 2021 (UTC)

Σ(1919)

A few things about this result:

1. Does the independence of S(1919) also apply to Σ(1919)?
2. Due to [2], is ZFC able to prove that the 1919-state busy beaver machine halts at all? (Scott Aaronson's original 7918-state TM can't be proven to halt by Godel's 2nd incompleteness theorem)
3. In the case that the first question is true, should the result in the article be improved to Σ(748)? C7X (talk) 00:44, 14 February 2021 (UTC)

1. I first thought that it is correct, but I made a mistake. So I have no idea.

2. Although I am too lazy to check the link, ZFC is not able to prove the termination, if I correctly understand. Does the paper include a result implying the termination?

3. I think so (if we assume the correctness of the reasoning of 1 and the paper, of course). But I do not know whether such a implication in 1 holds or not.

p-adic 02:24, 14 February 2021 (UTC)

2. "...construct a 7918-state machine which halts only if ZFC is inconsistent...Unpublished work by Stefan O’Rear brought this limit down to around 1900 states". From this wording I assume so.

3. Edit: Apparently #1 isn't needed and the result proven was "There’s an explicit 748-state Turing machine that halts iff ZF is inconsistent. Thus, assuming ZF is consistent, ZF cannot prove the value of BB(748)." (O'Rear) C7X (talk) 03:11, 14 February 2021 (UTC)

Thank you. Now I checked the paper, but what are written are weird for me.

First, it says "Specifically, they construct a 7918-state machine which halts only if ZFC is inconsistent: to prove that it halts would be to prove ZFC inconsistent," Since they are working in ZFC set theory, it means ZFC|-(Halt(X)→¬Con(ZFC)), where X is the TM. It does not mean ZFC+Con(ZFC)|-¬(ZFC|-Halt(X)), because ZFC+Con(ZFC) does not formally imply ¬(ZFC|-¬Con(ZFC)).

Secondly, it says "but to prove in ZFC that it does not would also prove that ZFC is inconsistent, via Godel's second incompleteness theorem." It means ZFC|-(ZFC|-¬Halt(X))→¬Con(ZFC)), but the deduction is wrong that Halt(X)→¬Con(ZFC), which is the only written property of X, does not imply "¬Halt(X)→Con(X)". Moreover, even i ZFC|-(ZFC|-¬Halt(X))→¬Con(ZFC)) is correct by some unwritten reasoning, it does not imply ZFC+Con(ZFC)|-¬(ZFC|-¬Halt(X)), again because ZFC+Con(ZFC) does not formally imply ¬(ZFC|-¬Con(ZFC))

Therefore the written discussion in the paper to deduce the undecidability of the halting problem looks wrong. (If it is written by a professional mathematician, then maybe I am wrong. Do you know the author? Please tell me whether the author is an actual mathematician or not.)

p-adic 06:23, 14 February 2021 (UTC)

Should the formula be "ZFC|-(ZFC|-Halt(X))→¬Con(ZFC))" in your message? I don't know the author, and apparently the paper uses BB to mean "maximum shifts function" and Σ to mean Rado's Σ function, so the results of Σ(748) and BB(748) aren't necessarily equivalent C7X (talk) 06:47, 14 February 2021 (UTC)

> Should the formula be "ZFC|-(ZFC|-Halt(X))→¬Con(ZFC))" in your message?

Your formula has an extra ")", but I guess that it should not be. Which formula in my message are you talking about?

p-adic 06:53, 14 February 2021 (UTC)

""but to prove in ZFC that it does not would also prove that ZFC is inconsistent, via Godel's second incompleteness theorem." It means [this formula] ZFC|-(ZFC|-Halt(X))→¬Con(ZFC)" (I think your quote also had an extra parenthesis) C7X (talk) 17:32, 14 February 2021 (UTC)

I see. But the original sentence in the paper says "to prove in ZFC that it does not", and a previous sentence says "to prove that it halts". Therefore I guessed that it meant "to prove in ZFC that it does not halt". That is why I wrote ZFC|-(ZFC|-¬Halt(X))→¬Con(ZFC), which I explained to be based on an illogical deduction by the author.

p-adic 04:01, 15 February 2021 (UTC)

OK.

> It does not mean ZFC+Con(ZFC)|-¬(ZFC|-Halt(X))

If this is the indended result, isn't ZFC+Con(ZFC)|-a true for any assertion a by Godel's 2nd incompleteness theorem, and by the principle of explosion? Here I am assuming that the "Con(ZFC)" is an assertion not in the meta-theory, but in the theory, and a is an assertion in the coded theory

> the deduction is wrong that Halt(X)→¬Con(ZFC), which is the only written property of X, does not imply "¬Halt(X)→Con(X)"

(From what I understand unless I'm making a mistake) This would be true since Halt(X) if and only if ZFC is consistent, instead of a one-way implication

> "machine which halts only if ZFC is inconsistent: to prove that it halts would be to prove ZFC inconsistent" ... ZFC|-(Halt(X)→¬Con(ZFC))

Is another error here that ZFC|-(Halt(X)→¬Con(ZFC)) doesn't necessarily imply (ZFC|-Halt(X))→(ZFC|-¬Con(ZFC)) (the second formula is formulated in this way because we're working in ZFC, i.e. from "proving [in ZFC] it halts would be proving [in ZFC] that ZFC is inconsistent")? C7X (talk) 05:08, 15 February 2021 (UTC)

> If this is the indended result, isn't ZFC+Con(ZFC)|-a true for any assertion a by Godel's 2nd incompleteness theorem, and by the principle of explosion?

Could you tell me how you applied Goedel's theorem?

> This would be true since Halt(X) if and only if ZFC is consistent, instead of a one-way implication

My point is that the author just asserted the one-way implication in the reasoning. (Also, I guess that "consistent" in your comment is perhaps a typo of "inconsistent", as the author asserts so.)

> ZFC|-(Halt(X)→¬Con(ZFC)) doesn't necessarily imply (ZFC|-Halt(X))→(ZFC|-¬Con(ZFC))

But ZFC|-(Halt(X)→¬Con(ZFC)) implies (ZFC|-Halt(X))→(ZFC|-¬Con(ZFC)) by →-elimintion.

p-adic 06:05, 15 February 2021 (UTC)

5-state record holder

The 5-state record holding machine seems to repeatedly apply the function f(x) = x+2round(x/3)+4. But why does it halt, and why is 4098 two off from a power of 2? A Hippopotatomus (talk) 04:26, 17 February 2021 (UTC)

Busy Beaver (16)

Daniel Nagaj (http://www.quantum.physics.sk/people/nagaj/) has built a Turing machine (16 states) that beats the Graham number. here is the link http://morphett.info/turing/turing.html?197640ce0f380f8a6b0a4cdd138156a0 can someone check: 1.Does the program work correctly? 2. Which lower bound for this machine is greater than the Graham number? Konkhra (talk) 00:46, 21 December 2021 (UTC)

How did you know that the code was written by Daniel Nagaj? 🐟 Fish fish fish ... 🐠 02:10, 21 December 2021 (UTC)

I didn't ask a question about the author of the code. but if it is so important to you. read the top lines of the code carefully. there it is clearly written who the author of the code is. and I personally wrote to him. and he answered me.Konkhra (talk) 02:41, 21 December 2021 (UTC)

I see. Thank you. 🐟 Fish fish fish ... 🐠 03:15, 21 December 2021 (UTC)

Total number of Turing machines

The article says: \(\Sigma(n)\) also can be defined as the largest number in set \(T = \{n_1,n_2,\cdots,n_k\}\), which contains outputs of all Turing machines with \(n\) states and 2-colors. The size of \(T\) is not larger than \((4n+4)^{2n}\) (total number of Turing machines with \(n\) states).

But how much set \(T = \{n_1,n_2,\cdots,n_k\}\) for 3-colors or 4-colors? what is the formula for counting total number of Turing machines with \(n\) states and \(m\)colors? Konkhra (talk) 06:51, 3 February 2022 (UTC)

If I correctly understand the question, it is bounded by 2^{m-1}×(2nm)^{nm}. (It depends on the formulation a little.)

p-adic 07:01, 3 February 2022 (UTC)

Thank you.Konkhra (talk) 07:06, 3 February 2022 (UTC)

\(\Sigma(6)>10\uparrow\uparrow 15\)

https://www.sligocki.com/2022/06/21/bb-6-2-t15.html

LittlePeng9 (talk) 10:48, 23 June 2022 (UTC)

I guess Sigma (8) will beat the graham number. Again, I guess. Sigma 15 at the level of f epsilon in fgh. And so sigma 50 absolutely crushes the loader number Konkhra (talk) 11:47, 23 June 2022 (UTC)

Comparision?

I doubt the comparison at BB(6) estimated. The estimation seems smaller than 10^^15. It should be like 10^^9. Also need updating the below parts at BB(7) —Preceding unsigned comment added by Octogeddonling (talk • contribs)

Source is required. 🐟 Fish fish fish ... 🐠 16:28, 18 May 2024 (UTC)

How large will Σ(54,42) be?

Is Σ(54,42)(54-state, 42-symbol) larger than GGGG...GGGG64 with G64 G's? Mika2005CN (talk) 08:01, 20 October 2024 (UTC)

Much more. See: https://wiki.bbchallenge.org/wiki/Champions. and this is a very, very conservative estimate. Konkhra (talk) 04:42, 29 October 2024 (UTC)

All n-state, m-sign Turing machines have corresponding m x ceil(log2(n))-state, 2-symbol equivalent Turing machines, so Σ(270)<Σ(54,42)<Σ(324). But I'm curious whether Σ(270) will be larger than GGGG... GGGG64 with G64 G's and TREE(3) or not. Mika2005CN (talk) 19:24, 26 October 2024 (UTC)

I don't know English well enough to explain it to you in detail. But there are indirect indications that Σ(80,2) is greater than growth rate PTO (Z_2), where Z_2 - second-order arithmetic. Konkhra (talk) 04:47, 29 October 2024 (UTC)

I recommend you to take an interest in for example binary lambda calculus (BLC). A function of 415 bits (https://github.com/tromp/AIT/blob/master/fast_growing_and_conjectures/bms.lam), written in this language, has a growth rate of PTO (Z_2). and Σ(80,2) is somewhere around 1300 bits. yes, BLC is growing faster than bb, but I believe, bb(80,2) beats growth rate of PTO (Z_2) and absolutely any well-defined ordinal notation today.Konkhra (talk) 04:52, 29 October 2024 (UTC)

Oh, I realized that ALL n-state, m-sign Turing machines have corresponding m x ceil(log2(n))-state, 2-symbol equivalent Turing machines, but NOT ALL m x 2^n-state, 2-symbol have corresponding n-state, m-sign Turing machines, maybe Σ(54,42) or Σ(270), which is bigger is uncertain. But Certainly Σ(54)<Σ(54,42)<Σ(324).

Mika2005CN (talk) 14:55, 29 October 2024 (UTC)

And, will Σ(54) larger than TREE(3)? Will Σ(324) larger than Loader?

Mika2005CN (talk) 15:10, 29 October 2024 (UTC)

Do you understand how huge PTO (Z_2) is? TREE is simply indistinguishable from 0 even in comparison with a function with the growth rate of the Buchholz ordinal in FGH.Konkhra (talk) 00:42, 30 October 2024 (UTC)

Loader is very big. in BLC language it takes 2000 bits (https://codegolf.stackexchange.com/a/274634). So i don't believe Σ(324) will beat it.Konkhra (talk) 00:46, 30 October 2024 (UTC)

but overall there are no answers to your questions. my answers are rough guesses.Konkhra (talk) 06:39, 30 October 2024 (UTC)