Talk:Busy beaver function/Archive 1

Is \(f_{\omega^\text{CK}_1}(n)\) a well-defined function? How to evaluate \(f_{\omega^\text{CK}_1}(2)\), for example? Ikosarakt1 (talk) 11:52, January 28, 2013 (UTC)


 * We can't, it's uncomputable xD
 * I have no idea what the fundamental sequence for \(\omega^\text{CK}_1\) is &mdash; probably different functions of \(\omega\) stepping up through different levels of recursion. That is, if a fundamental sequence exists. FB100Z &bull; talk &bull; contribs 20:40, January 30, 2013 (UTC)


 * Or maybe \(f_{\omega^\text{CK}_1}\) will have to be defined backwards from \(\Sigma\). I dunno. FB100Z &bull; talk &bull; contribs 21:39, February 1, 2013 (UTC)

I found site which conjectures definition of \(f_{\omega^\text{CK}_1}(n)\). If it is really that, \(f_{\omega^\text{CK}_1}(2) = f_{\omega}(2) = 2^2 \times 2 = 8\). Ikosarakt1 (talk) 09:05, February 2, 2013 (UTC)

\(\Sigma(5)\)
Probably, we soon find out whether \(\Sigma(5) = 4098\) or not. According to this list, only 35 machines remains undecided. Ikosarakt1 (talk) 10:07, February 22, 2013 (UTC)

Unfortunately, that list is over a decade old and no one has really made progress since. This doesn't seem to be the sort of problem that interests many mathematicians. (There is one J Harland who mentioned he was working towards solving this, but again that was back in 2009.)Deedlit11 (talk) 15:00, February 28, 2013 (UTC)

Mathematicians don't care about us *sob* FB100Z &bull; talk &bull; contribs 07:10, March 5, 2013 (UTC)

Grounding
Nice explanation, FB100Z. I want more clouds! 02:51, March 5, 2013 (UTC)

That was my work, read history. Ikosarakt1 (talk) 06:29, March 5, 2013 (UTC)
 * Aw, you ruined my moment of stolen glory :P FB100Z &bull; talk &bull; contribs 07:04, March 5, 2013 (UTC)


 * Oops! I didn't notice that. Anyway, thanks for improving the explanation, FB100Z! &mdash; I want more clouds! 08:22, March 5, 2013 (UTC)

I'm going to add oracle Turing machines to the informal explanation, but I haven't found a definition that makes sense and is reasonably concrete. Anyone wanna help? FB100Z &bull; talk &bull; contribs 18:47, March 5, 2013 (UTC)

I believe that oracle Turing machine looks like that: we have two tapes, two rulesets and oracle box: normal and oracle. Then oracle box can take arbitrary Turing machine program (which is prescribed in the oracle ruleset), and test whether it halts or not. If not, then write nothing at the current oracle cell blank or write 1 otherwise, move into the direction according to the oracle ruleset. Ikosarakt1 (talk) 21:06, March 5, 2013 (UTC)

Nice construction, Ikosarakt. I had conceived of an oracle as something like the following:  it's a Turing machine with one tape and one ruleset, however, there is a special oracle state (we'll call it state O). When in the oracle state, the Turing machine does not check whether the currect cell is 1 or 0, but rather counts the number of consecutive 1's starting from the currect cell and moving to the left. If that number is n, the oracle will instantly check whether the nth Turing machine halts or doesn't halt. If it halts, one set of instructions (go to state, write either 0 or 1, move either right or left) is executed;  if it doesn't halt, another set of instructions is executed. I think this is powerful enough to define an oracle machine, although I haven't proven it. Deedlit11 (talk) 20:44, March 6, 2013 (UTC)

Number of consecutive 1's can be very large and larger than the number of TM's with some number of states. For example, there exists 20736 TM's with 2 states, but if there will be, say, 30000 consecutive 1's, we need to check non-existed TM, thus the function freeze. Ikosarakt1 (talk) 22:24, March 6, 2013 (UTC)

The oracle in my oracle Turing machine is not limited to 2-state or n-state Turing machines;  it can test a Turing machine with any number of states (limited of course by the maximum number of 1's the oracle Turing machine can print out). It's actually important that the oracle can test Turing machines with arbitrarily many states. If your n-state oracle Turing machine can only test Turing machines with n states or less, than it can't evaluate BB(n+1), for example. So it can't evaluate the general function BB(m), much less BB^m(m) and higher functions. So your oracle Turing machine is far less powerful than it needs to be.

Remember, we need to be able to evaluate any algorithm where functions Halt(m) and BB(m) are allowed. Deedlit11 (talk) 23:17, March 6, 2013 (UTC)

Multi-head TMs
Let reconstruct TM such that head moves instead of tape. Then we can add more heads and for each head specify a separate ruleset. Initially all heads occupy one position. Define \(\Sigma(a,b,c)\) to be largest finite number of non-zeros that can be written on the a-state, b-color, c-head Turing machine. Ikosarakt1 (talk ^ contribs) 10:52, March 15, 2013 (UTC)
 * Since a busy beaver uses up a finite amount of space, we can place the machines far apart from each other so they don't interact during their computations. This means that in general \(\Sigma(a, b, c \cdot d) \geq \Sigma(a, b, c) \cdot d\). Also \(S(a, b, c) \geq S(a, b)\). FB100Z &bull; talk &bull; contribs 18:24, March 18, 2013 (UTC)

Yes, but under my definition, all heads initially started from the same place. Ikosarakt1 (talk ^ contribs) 18:31, March 18, 2013 (UTC)
 * Oops. I believe the bounds still hold, though. FB100Z &bull; talk &bull; contribs 18:47, March 18, 2013 (UTC)

That's an example of 1-state, 2-headed Turing machine that writes 2 ones on the tape and performs 3 steps. Nearest caret indicates the position of 1st head and otherwise.

Ruleset for the 1st head: a _ 1 r halt

Ruleset for the 2nd head: a _ 1 r halt a 1 1 l a

0000000000 1st head state = a; 2nd head state = a; Steps = 0 ^   ^

0000100000 1st head state = a; 2nd head state = halt; Steps = 1 ^

0000100000 1st head state = a; 2nd head state = halt; Steps = 2 ^

0000100000 1st head state = halt; 2nd head state = halt; Steps = 3

Therefore, \(\Sigma(1,2,2) \geq 2\) and \(S(1,2,2) \geq 3\). Ikosarakt1 (talk ^ contribs) 18:57, March 18, 2013 (UTC)

BB(50)
I know there is proof that the BB(64) greater than the Graham's number. can we prove that, for example BB (50) greater than the  Graham's number ? Konkhra ( talk ) 08:02, March 26, 2013 (UTC)

We can't prove it if it isn't true. Only obvious way to do it is to find explicit 50-state TM which writes more than G symbols. If there is no such machine, I'm pretty sure we win't be able to prove it. LittlePeng9 (talk) 14:45, March 26, 2013 (UTC)

I think the best way to find bounds for values of TMs is to develop a micro-language that can compile into Turing machines. Maybe find an isomorphism between, say, Brainfuck and TMs? FB100Z &bull; talk &bull; contribs 17:34, March 26, 2013 (UTC)


 * That may make it easier to write programs that right large numbers, at the expense of parsimony; running a program through an interpreter won't get us Turing machines as small as if we programmed Turing machines natively. The question is if it is really easier to program larger numbers in Brainfuck than in Turing machines. How short a Brainfuck program can we write that outputs a number larger than Graham's number? Deedlit11 (talk) 23:31, March 27, 2013 (UTC)

Sorry for a silly question, but what is an isomorphism? Ikosarakt1 (talk ^ contribs) 10:10, March 27, 2013 (UTC)

Here I think we can define it as strong correlation. Compiler of TM in Brainfuck would allow us to see how is BB_F(n) (busy beaver function for Brainfuck) related to BB(n) LittlePeng9 (talk) 12:48, March 27, 2013 (UTC)

"Don't move" variation
How about variation of \(\Sigma(n)\) such that TMs also allowed to don't move (just change state) in their rules? Ikosarakt1 (talk ^ contribs) 11:39, April 2, 2013 (UTC)

No improvement. Consider machine which, when entering square with symbol x in state s, doesn't move for a while. When it finally leaves, it moves, say, to the right, leaves x' on tape and transites to state s'. Same effect would be achived with single transition s,x->x',s',r. So it can be trivially emualted without increasing number of states. It does affect S(n), however, but not significantly. LittlePeng9 (talk) 11:55, April 2, 2013 (UTC)

So there exists constant c such that \(\Sigma_{\text{don't move allowed}}(n) > \Sigma(n+c)\). Good, what is this constant? Ikosarakt1 (talk ^ contribs) 12:19, April 2, 2013 (UTC)

While you ask for Sigma function, c=0. For max shift function, for suffisciently large n, \(S_{\text{don't move allowed}}(n)>S(n+1)\). LittlePeng9 (talk) 12:51, April 2, 2013 (UTC)


 * Without introducing oracle TMs, there's not much we can do to escape the Church-Turing thesis. FB100Z &bull; talk &bull; contribs 16:51, April 2, 2013 (UTC)

First, there was no attempt to create function that grow much faster than normal \(\Sigma(n)\), it's just a variation.

Second, how to create functions that fill intermediate growth rates between \(f_{\omega^\text{CK}_1}(n)\) and \(f_{\omega^\text{CK}_2}(n)\)? Ikosarakt1 (talk ^ contribs) 20:05, April 2, 2013 (UTC)
 * \(\omega_1^\text{CK} + \) any large recursive ordinal? FB100Z &bull; talk &bull; contribs 21:45, April 2, 2013 (UTC)

Not quite. I mean the ordinals something like \({\omega_1^\text{CK}}^{\omega_1^\text{CK}}\) or, say, \(\varepsilon_{\omega_1^\text{CK}+1}\). In other words, the ordinal recursion around \(\omega_1^\text{CK}\), but still smaller than \(\omega_2^\text{CK}\) Ikosarakt1 (talk ^ contribs) 22:08, April 2, 2013 (UTC)

Well, you can always take the FGH at those ordinals - but I guess you want something other than an ordinal hierarchy. Deedlit11 (talk) 07:51, April 3, 2013 (UTC)

By the way, when don't move allowed, then \(S(1) = 2\), using rules 0 _ 1 * 0 and 0 1 * r halt. Ikosarakt1 (talk ^ contribs) 15:47, April 20, 2013 (UTC)

There was no typo. Let D in subscript mean don't move variation. Halting machine spends only finite time in cell and, when entering with same state and symbol, always leaves same (possibly different from first one) symbol. So set of transitions in single cell can be replaced with single move transition. Call this new machine a cut-machine. Cut-machine leaves exactly same output, so \(\Sigma(n)=\Sigma_D(n)\).

Let's look at how much does cut-machine really cut. Normal machine can stay in single cell only finite time. If it loops, it must stay there forever. From Dirichlet's principle it can stay there only 2n steps (number of state-symbol combinations). So machine can make at most 2n times more steps than its cut-machine. Cut-machines are standard TMs, so we have bound \(S_D(n)\leq 2n\times S(n)\). Because S(n) is uncomputable, \({S(n+1)}\over S{(n)}\) grows uncomputably fast, so, for suffiscently large n, \({S(n+1) \over S(n)}>2n\), i.e. \(2n\times S(n)2 if we note that due to halt condition 2n can be replaced with 2n-1. LittlePeng9 (talk) 17:20, April 20, 2013 (UTC)

Unfortunately, while \(\lfloor \frac{S(n+1)}{S(n)} \rfloor\) is necessarily uncomputable, we don't know that it is particularly large at any particular value, nor that it is eventually large, only that it is large infinitely often. So we don't know that \(2n \times S(n) < S(n+1)\) for sufficiently large n, only for infinitely many n. Not that I doubt it is true, unfortunately we can't even prove weak statements like \(\Sigma(n+1) > \Sigma(n)\) for sufficiently large n. Deedlit11 (talk) 17:33, April 20, 2013 (UTC)

Unsurprisingly, we can prove \(\Sigma(n+1) > \Sigma(n)\). Consider n-state busy beaver. We replace its halt transition with "move to (n+1)th state". This state moves right without altering nonblank symbols, and when it hits blank it halts leaving 1. Thus \(\Sigma(n+1)\geq \Sigma(n)+1\) and \(\Sigma(n+1) > \Sigma(n)\). This proof that \(\Sigma(n)\) is strictly increasing is somewhat elementary. I think I've seen once proof that \(S(n+1)-S(n)\) is strictly monotonic, although I'm not sure. LittlePeng9 (talk) 18:08, April 20, 2013 (UTC)

Whoops, I messed that up! Replace it with \(\Sigma(n+1) > 2 \Sigma(n)\) for sufficiently large n then. It's very hard to prove similar statements even though they seem obvious. Deedlit11 (talk) 18:25, April 20, 2013 (UTC)

You are right that such inequality can be hard to prove. I think that general statement, while very hard to prove, is true: for every computable function \(f: \mathbb{N} \rightarrow \mathbb{N}\) inequalities \(f(\Sigma(n))<\Sigma(n+1)\) and \(f(S(n))<S(n+1)\) work for all suffiscently large \(n\). Note that this doesn't hold for all strictly increasing uncomputable functions. LittlePeng9 (talk) 14:12, May 9, 2013 (UTC)

Uncomputability of \(f(n)=\Sigma(c,n)\)
Is \(f(n)=\Sigma(c,n)\) is computable? Here c means constant, for example \(f(n)=\Sigma(10,n)\). In other words if number of states remains unchanged and number of colors changes, will this function be computable? Ikosarakt1 (talk ^ contribs) 15:41, April 4, 2013 (UTC)

Of course \(\Sigma(1,n)\) is computable. I'm pretty sure that \(\Sigma(c,n)\) is uncomputable for \(c \ge 2\). In fact I believe that you can compute arbitrary partial recursive functions with two states and arbitrarily many colors. For example, there is a 2-state, 18-symbol universal Turing machine. I'm not quite sure how to go from there to arbitrary partial functions on a blank tape, but I believe it follows. Deedlit11 (talk) 03:44, April 5, 2013 (UTC)

If I understand correcly, the same must be applied to \(f(n) = \Sigma(c_1,c_2,n)\), \(f(n) = \Sigma(c_1,c_2,c_3,n)\)? Here 3rd argument responds to the number of heads, and 4rd for the number of dimensions. Ikosarakt1 (talk ^ contribs) 13:48, April 5, 2013 (UTC)

Hmmm, it would help to have Turing machines defined explicitly for multiple heads and multiple dimensions - exactly what is read, what instructions are allowed, do we move all the heads or just one head, do we move in one dimension or multiple dimensions, etc.

In any case, I believe that \(f(n) = \Sigma(c_1,c_2,c_3,n)\) is computable, in fact it is eventually constant. The reason is that with \(c_1, c_2, c_3\) fixed, there are a fixed number of instructions per machine, so only a fixed number of dimensions can be utilized. Any additional dimensions are just wasted.

As for \(f(n) = \Sigma(c_1,c_2,n)\), I need the exact specification for multiple heads. According to

http://www.cs.odu.edu/~toida/nerzic/390teched/tm/othertms.html

and

http://www.cs.usfca.edu/~sjengle/ucdavis/ecs120-f06/doc/discussion-111506.pdf

each state can utilize only one head. So there is no need to have more heads than states, so \(f(n) = \Sigma(c_1,c_2,n)\) will be eventually constant again.

The question becomes more interesting if one is allowed to take input from all the heads. If I had to guess, I would guess that you generally get Turing completeness, perhaps even for (2, 2, n). Deedlit11 (talk) 00:55, April 6, 2013 (UTC)

For multihead TMs, each head has specified set of rules (transition table) and they initially start from the same cell. I've managed the example of 2-head TM above. I'm not sure that it is eventually constant, since each head changes the behaviour of the other.

For multidimensional TMs, I believe that we can create "coordinate" row to define transitions in higher dimensions. For example, (l,r,l,r) means that "go to the left one cell in the first dimension, then to the right cell in the second dimension, then to the left cell in the third dimension, then to the right cell in the fourth dimension". You may ask what exactly means transition "to the right cell in the second dimension": down or up. However, it is not necessary to bother about it. (l,r,l,r) means that coordinates changed from (a,b,c,d) to (a-1,b+1,c-1,d+1). For instance (0,0,0,0) changes to (-1,1,-1,1). That defines explicitly transitions between dimensions.

Therefore, I believe that the rule for n-dimensional TM might look like:



Ikosarakt1 (talk ^ contribs) 10:12, April 6, 2013 (UTC)

For multidimensional TMs: Then \(f(n) = \Sigma(c_1,c_2,c_3,n)\) will indeed be eventually constant, for the reasons I explained above.

For multihead TMs: Ah, so each head gets its own set of instructions? Interesting. Do you cycle through the instruction sets, applying the rule for the first head, then the rule for the second head, etc.? It appears that this is somewhat weaker than having a single instruction set that reads all inputs, but I wouldn't be surprised if you still got Turing completeness for (c_1, c_2, n). I'm not sure though. Deedlit11 (talk) 03:04, April 8, 2013 (UTC)

Yes, I meant that construction. Ikosarakt1 (talk ^ contribs) 11:31, April 10, 2013 (UTC)

I want to officially announce that \(\Sigma(c,n)\) is uncomputable. I'll sketch a proof here. First of all, we need Universal Turing Machine V with following property: for every halting problem we can construct input such that machine in question halts iff V halts on given input. We can create arbitrary input with bounded number of states using construction similar to this: 0 _ a r 1 0 b b r 1 0 c c r 1' 0 d d r 1 0 e e r 1 1 _ 1 l 2 1' _ 0 l 2 2 a b l 3 2 b c l 3 2 c d l 3 2 d _ l 3 2 e _ r halt 2 _ _ l 0 3 _ e r 2 3 e d r 2 3 d c r 2 3 c b r 2 For every character of input we have one color. This example writes binary expansion of number 27. Now we can have fixed machine which translates number n written in binary to input for V corresponding to nth halting problem. Finally we use V to "solve" this halting problem. Call resulting machine M.

Sketch of proof that this construction implies uncomputability - if we can compute \(\Sigma(c,n)\) we can give upper bound on \(S(c,n)\). By analysing how many colors we need to solve given halting problem and using that number together with state count of M we'd know how many states M can execute. So by running M for \(S(c,n)\) steps gives answer to given halting problem. Contradiction. LittlePeng9 (talk) 16:54, May 3, 2013 (UTC)

So, this assumes the existence of a machine that, given a number n as input, outputs the correct input for V corresponding to the nth halting problem (Note: you don't want n written in binary in the usual way, as Turing machines can't read an arbitrary binary number.). Do we know that such a machine exists? I think we would need to know the specifics of say the particular UTM. Deedlit11 (talk) 00:40, May 4, 2013 (UTC)

Number theory
Is \(\Sigma(5)\) even or odd? What is the last digit of \(\Sigma(7)\)? Will we ever know? FB100Z &bull; talk &bull; contribs 20:04, April 14, 2013 (UTC)

To determine numerical properties of \(\Sigma(n)\), we should know TM that computes it. As far as we don't know it, we can't determite it. By the way, there are relevant topic. Ikosarakt1 (talk ^ contribs) 20:25, April 14, 2013 (UTC)

It may be that \(\Sigma(5)\) is 4098, and someone may be able to prove it in the not to distant future. I imagine that \(\Sigma(7)\) will never be found, and I don't see how we could ever find the last digit without finding the number itself. Deedlit11 (talk) 00:02, April 17, 2013 (UTC)

I hope the quantum computer will help us to find \(\Sigma(7)\). May be in the middle of the 21st century or even earlier.Konkhra (talk) 00:06, April 17, 2013 (UTC)

Due to speed and memory of quantum computers will grow hyper-exponentially (something like \(2^{2^n}\) where n increases linearly after each couple of years), we can assume that, say, in year 2050 mankind will have computer with \(10^{1000000}\) operations per second. Therefore, it will be helpful for finding not only \(\Sigma(7)\), but much more larger values: \(\Sigma(100)\), \(\Sigma(1000)\), and so on. Although they will not compute the exact values, but they will able at least weed out small TM's, which sufficiently makes analysis easier. Ikosarakt1 (talk ^ contribs) 10:01, April 17, 2013 (UTC)

I think the popular media has left people with the impression that quantum computing will lead to speed and memory increasing exponentially with the number of qubits. However, this is a misconception. It is true that n qubits, when in superposition, require \(2^n\) complex numbers to specify. However, when you try to make any measurement, the system decoheres and you wind up with only one result. So you don't get the power of a \(2^n\) bit computer. This is still good enough to make a narrow class of problems faster, including factoring, database search, and simulation of quantum systems. But for most problems there does not appear to be an exponential improvement.

In any case, I don't think weeding out small TM's for 100 state and 1000 state Busy Beavers will help much. Even for \(\Sigma(6)\) and \(\Sigma(7)\), there will be thousands of "chaotic" machines where determination of halting is very difficult, and I think this is the hardest part of the Busy Beaver problem. Deedlit11 (talk) 16:36, April 20, 2013 (UTC)

Yes, finding solution of halting problem isn't easy problem even for Σ(5). For example try to run something like TM#827123 for 5 states:

0 _ 1 l 2 0 1 1 l 4 1 _ 1 l halt 1 1 1 l 3 2 _ 1 r 3 2 1 _ l 3 3 _ 1 l 0 3 1 1 r 4 4 _ _ l 1 4 1 _ r 2

It is hard to find any periodic for this TM, although I believe that some enthusiast(s) eventually will discover this. Ikosarakt1 (talk ^ contribs) 17:59, April 20, 2013 (UTC)

Number of BB's
Can we determine number of Busy Beavers with n states without knowing the table of each? Ikosarakt1 (talk ^ contribs) 17:12, April 16, 2013 (UTC)


 * I doubt it. You need to find the Busy Beavers to know how many there are. We know the number is divisible by 4 though, since we can switch left and right, and the halting state can be either left or right. Deedlit11 (talk) 20:04, April 16, 2013 (UTC)

Reverse BB
Suppose a Turing machine starting with a blank tape outputs exactly n 1s. What's the minimum number of states needed? FB100Z &bull; talk &bull; contribs 01:36, April 21, 2013 (UTC)

This problem is form of complexity measure. Let K(n) be answer to your question. By Chaitin's incompleteness theorem for every mathematical theory (like ZFC) there exists number a such that no true sentence of form K(n)>a is provable within that theory. For ZFC a<10^^10. LittlePeng9 (talk) 05:52, April 21, 2013 (UTC)

James Harland has studied these machines, and calls them Placid Platypuses. (Platypi?)  He wrote a paper here. Deedlit11 (talk) 06:18, April 21, 2013 (UTC)
 * I like the animal names :P FB100Z &bull; talk &bull; contribs 06:28, April 23, 2013 (UTC)

QTM
So...quantum Turing machines? FB100Z &bull; talk &bull; contribs 06:52, April 23, 2013 (UTC)


 * I don't know much about quantum Turing machines, but I do know that quantum computers can be simulated by regular Turing machines, so the quantum Busy Beaver function will not grow fundamentally faster than the regular Busy Beaver function. Deedlit11 (talk) 15:08, April 27, 2013 (UTC)


 * But this page says that Turing and Church did not prove that quantum computers are equivalent to Turing machines! -- I want more clouds! 15:14, April 27, 2013 (UTC)


 * I don't think Munafo is an authoritative source. I've certainly read that quantum computers can be simulated by classical computers in many places, for example here. Munafo is also incorrect that nondeterministic computers are not emulatable by deterministic computers, although he may be referring to a different definition of nondeterministic. Deedlit11 (talk) 15:56, April 27, 2013 (UTC)


 * No matter which definition he had in mind, most, if not all, nondeterministic TM-like systems are simulateable by simultaneously checking all choice paths, or checking them sequentially. LittlePeng9 (talk) 16:45, April 27, 2013 (UTC)

2 state, 5 color BB-contender
Anyone have the info about final condition of the tape of this TM? Knowing this, I think I can prove that \(\Sigma(6,5) > 10 \uparrow\uparrow (1.7*10^{352})\). Ikosarakt1 (talk ^ contribs) 13:36, May 9, 2013 (UTC)

Wait, after some analysis I figured out that it looks like:

142222...2222 (~1.7*10^352 2's)  ^

Is it indeed true? Ikosarakt1 (talk ^ contribs) 13:48, May 9, 2013 (UTC)

Number of halting TMs
Using exact values of \(\Sigma(n)\) for n<5 and computer program I found that:

For 1 state, there will be 32 (50% of all) halting TMs.

For 2 states, there will be 9760 (~47.07% of all) halting TMs.

For 3 states, there will be 7521168 (~44.83% of all) halting TMs.

For 4 states, there will be 11007946176 (~43% of all) halting TMs.

If my computations are all correct, I can say that percent of halting TMs decreases when the number of states grows. Ikosarakt1 (talk ^ contribs) 19:59, May 13, 2013 (UTC)
 * Let \(h(n)\) be the number of halting TMs with \(n\) states. I conjecture that \(\lim_{n \rightarrow \infty} h(n)/(4n + 4)^{2n} = K\) for some positive constant \(K\). If this is correct, then \(K\) is probably related to . FB100Z &bull; talk &bull; contribs 23:31, May 13, 2013 (UTC)

Relationship between \(S(n,m)\) and \(\Sigma(n,m)\)
I guess that \(S(n,m) \approx \Sigma(n,m)^2\), considering different known values and bounds of both. Ikosarakt1 (talk ^ contribs) 12:39, May 24, 2013 (UTC)


 * That's the standard growth rate for "Christmas trees". Turing machines like counters have relationships between \(S(n,m)\) and \(\Sigma(n,m)\) Deedlit11 (talk) 10:31, May 25, 2013 (UTC)
 * But why almost all record holders have that relationship? Ikosarakt1 (talk ^ contribs) 11:04, May 25, 2013 (UTC)

Is the second entry is the 'amount of nested'? $Jiawhein$\(a\)\(l\)\(t\) 10:54, May 25, 2013 (UTC)
 * No, it is the number of types of symbols that Turing machine can use. Ikosarakt1 (talk ^ contribs) 11:04, May 25, 2013 (UTC)