Talk:Busy beaver

message: I propose the following introduction:
: The Busy Beaver Game consists of desiging a halting, binary-alphabet Turing Machine which writes the most 1s on the tape, using only a limited set of states. The rules for the 2-state game are as follows: (i) the machine must have two states in addition to the halting state, and (ii) the tape starts with 0s only. As the player, you should conceive each state aiming for the maximum output of 1s on the tape while making sure the machine will halt eventually.
: The Nth busy beaver or BB-n is the Turing Machine that wins the N-state Busy Beaver Game. That is, it attains the maximum number of 1s among all other possible N-state competing Turing Machines. The BB-2 turing machine, for instance, achieves four 1s in six steps.
: The busy beaver game has implications in computability theory, the halting problem, and complexity theory. The concept was first introduced by Tibor Radó in his 1962 paper, "On Non-Computable Functions". Negrulio (talk) 15:51, 27 December 2015 (UTC)
thread_title:  Proposed new introduction 

message: At the moment, only Σ (number of nonzeros in result) and S (number of steps taken) are mentioned.  There are others that could be investigated:
* number of symbol toggles
* number of toggles from 0 to 1 (or 1 to 0, 0 to non-0, to higher-index symbol, etc.)
* number of tape positions that are ever toggled
* final distance from starting point
* greatest distance reached from starting point
* greatest distance from starting point at which a non-0 is ever set
* greatest distance from starting point at which a non-0 is present at the end
* overall explored range
* overall range of non-0s ever set
* overall range of non-0s at the end
.... -- Smjg 15:11, 27 September 2005 (UTC)
thread_title:  Busy beaver functions 

message: I happened upon the Sci Am refrences in an old notebook of mine; they are "unverified" in that I haven't laid eyes on them in 20-odd years. The Rado reference actually appears in Booth's references for his chapter 9 titled "Turing Machines". So it is "quasi-unverified" as well. I'm staring at the pages of the Booth book, so I guess we could argue that reference is verified.
wvbailey22:08, 6 January 2006 (UTC)
replies: text: Or should it be https://homepages.hass.rpi.edu/heuveb/Research/BB/status.html ? 199.203.204.197 (talk) 06:55, 28 June 2022 (UTC)
thread_title:  References 

message: The pdf is currently available at this URL:
https://www.cs.auckland.ac.nz/~chaitin/bellcom.pdf
I don't know how to properly edit references, sorry.
68.7.137.69 (talk) 04:32, 26 May 2016 (UTC)
replies: text: Or should it be https://homepages.hass.rpi.edu/heuveb/Research/BB/status.html ? 199.203.204.197 (talk) 06:55, 28 June 2022 (UTC)
thread_title:  Link to Chaitin’s “Computing the Busy Beaver Function” is broken 

message: ==='1s' versus '1s' versus 'ones'===
The article makes inconsistent use of "1's", "1s", and "ones" to refer to the plural of "1". How should this be standardized? Also, what is meant by the notation "SRTM(n)"? Pmdboi 14:24, 8 March 2006 (UTC)
replies: text: Unless Wikipedia has a standard contrary to my opinion I vote for "1's" when the 1's are the plural of symbol "1" from the binary collection of symbols { 0, 1 }. Standard American prose/literature wants/uses/expects "one" and "ones". Interesting that the international symbol for OFF is O and ON is | (vertical-slash, not ordinal "1": per the IEC-- can't remember the exact specication number IEC-317, Symbols? ). Emil Post specified "mark" = { | } and "blank" = { } in his models. It probably should be "|", both plural and singular, but I'd stick with "1's" per the argument: "If Bill likes it, it must be good." (Hope this helps). wvbaileyWvbailey 00:45, 9 March 2006 (UTC)

text: As of December 2015 I can't find any "1's" or "0's". Everything has been replaced with "1s" and "0s". There is no English standard on how to write plurals of single digits. Oxford Dictionaries Online suggest adding no apostrophe to pluralize a number except when it is a single digit, in which case "you can use an apostrophe to show the plurals of single numbers." Notice the "can". Thus, this article is correctly written. Negrulio (talk) 14:17, 27 December 2015 (UTC)

text: ^{b}a = a\uparrow\uparrow b
 
But in the text we currently have a seemingly-unnecessary mixing of the two notations and a failure to identify the alternate form:
    \Sigma(2k) > 3 \uparrow^{k-2} 3 > A(k-2,k-2) \quad (k \ge 2),
    where \uparrow is Knuth up-arrow notation and A is Ackermann's function.
If we stick to just one notation, it should be equivalent to either:
    \Sigma(2k) > 3\uparrow(3\uparrow\uparrow(k-2)) > A(k-2,k-2) \quad (k \ge 2)
OR
    \Sigma(2k) > 3^{3\uparrow\uparrow(k-2))} > A(k-2,k-2) \quad (k \ge 2)
Finally, I find the use of a single \quad to be confusing (because it just looks like an extra space or so being added and still a multiply, as opposed to a clearer delimited) so I'd instead prefer something like:
    \Sigma(2k) > 3^{(3\uparrow\uparrow(k-2))} > A(k-2,k-2) \ where \ (k \ge 2)
I'll note that I'm not totally bent on this last modification if it's something that's really standard that I simply haven't seen or noticed before. Thanks. — Koyae (talk) 06:28, 25 December 2014 (UTC)

text: In case anyone was thinking of changing the article in response to this, please don't. There's no mixing of notations here; 3 \uparrow^{k-2} 3 means 3 \uparrow\cdots\uparrow 3, where k-2 is the number of arrows. 78.146.215.183 (talk) 15:52, 28 April 2015 (UTC)
thread_title:  Notation 

message: The article makes inconsistent use of "1's", "1s", and "ones" to refer to the plural of "1". How should this be standardized? Also, what is meant by the notation "SRTM(n)"? Pmdboi 14:24, 8 March 2006 (UTC)
replies: text: Unless Wikipedia has a standard contrary to my opinion I vote for "1's" when the 1's are the plural of symbol "1" from the binary collection of symbols { 0, 1 }. Standard American prose/literature wants/uses/expects "one" and "ones". Interesting that the international symbol for OFF is O and ON is | (vertical-slash, not ordinal "1": per the IEC-- can't remember the exact specication number IEC-317, Symbols? ). Emil Post specified "mark" = { | } and "blank" = { } in his models. It probably should be "|", both plural and singular, but I'd stick with "1's" per the argument: "If Bill likes it, it must be good." (Hope this helps). wvbaileyWvbailey 00:45, 9 March 2006 (UTC)

text: As of December 2015 I can't find any "1's" or "0's". Everything has been replaced with "1s" and "0s". There is no English standard on how to write plurals of single digits. Oxford Dictionaries Online suggest adding no apostrophe to pluralize a number except when it is a single digit, in which case "you can use an apostrophe to show the plurals of single numbers." Notice the "can". Thus, this article is correctly written. Negrulio (talk) 14:17, 27 December 2015 (UTC)
thread_title: '1s' versus '1s' versus 'ones'

message: In the statement about known values, there is a very strange usage of the reverse-superscript notation (which, if I understand correctly, connotes tetration). The reason I find it strange is that it's being mixed in with up-arrow notation which connotes the same thing. I find this to be a bad practice, as it should just be one or the other.
From the up-arrow notation article:
:
^{b}a = a\uparrow\uparrow b
 
But in the text we currently have a seemingly-unnecessary mixing of the two notations and a failure to identify the alternate form:
    \Sigma(2k) > 3 \uparrow^{k-2} 3 > A(k-2,k-2) \quad (k \ge 2),
    where \uparrow is Knuth up-arrow notation and A is Ackermann's function.
If we stick to just one notation, it should be equivalent to either:
    \Sigma(2k) > 3\uparrow(3\uparrow\uparrow(k-2)) > A(k-2,k-2) \quad (k \ge 2)
OR
    \Sigma(2k) > 3^{3\uparrow\uparrow(k-2))} > A(k-2,k-2) \quad (k \ge 2)
Finally, I find the use of a single \quad to be confusing (because it just looks like an extra space or so being added and still a multiply, as opposed to a clearer delimited) so I'd instead prefer something like:
    \Sigma(2k) > 3^{(3\uparrow\uparrow(k-2))} > A(k-2,k-2) \ where \ (k \ge 2)
I'll note that I'm not totally bent on this last modification if it's something that's really standard that I simply haven't seen or noticed before. Thanks. — Koyae (talk) 06:28, 25 December 2014 (UTC)
replies: text: In case anyone was thinking of changing the article in response to this, please don't. There's no mixing of notations here; 3 \uparrow^{k-2} 3 means 3 \uparrow\cdots\uparrow 3, where k-2 is the number of arrows. 78.146.215.183 (talk) 15:52, 28 April 2015 (UTC)
thread_title:  Use of up-arrow notation versus leading-superscript notation 

message: I have noticed a slight problem with this proof.  It assumes the number of states required for EvalS is constant no matter the value of n0.    What if the number of states required for EvalS varies according to n0?   Perhaps the turing machine encoding of EvalS can be computed from n0?
:No.  If there was a Turing machine M which could create EvalS based upon n0, then we could run M on n0 to create EvalS and then run EvalS, thus having a machine with fixed size which runs EvalS.  This is some justification for our definition of computability as Turing machine computability. - Sligocki 03:14, 9 December 2006 (UTC)
thread_title:  Error? 

message: Hi,
this article is pretty nice but it's very sophisticiated! I consider myself okay with maths and a minimum of computing but most of this goes over my head. Wikipedia should be accessible to laypeople, and even though it's hard when you're an expert and the audience is not, I think that the editors should dumb this down a little! If a middle school student who has no maths or computing skills cannot completely understand the article then it has no place in an encyclopedia! If something cannot be left out then include links as a minimum. -- ben 18:03, 11 July 2006 (UTC)
replies: text: Is there a category where we can put requests for accessibility? Im looking at this, and it seems that unless you have a background in whatever this article is talking about, you wont be able to understand it. for example, that sequence "1, 4, 6, 13, ≥ 4098 doesnt make any sense at all. It should definately be made more clear to the average person what the reasons are for not knowing what the fifth number is when it is known that it can't be bigger than 4098. Also, whats the point of all this? The article doesnt seem to make it clear why any of this is of any importance to anything, such as why anyone would ever want to know what the fifth value that is less than 4098 is, and how knowing that number would make any difference to anything. Definately not user friendyly Carterhawk 08:23, 26 August 2006 (UTC)
thread_title:  Accessibility 

message: thumbnail|500px|This is a two-symbol three-state Busy Beaver. It is working on a tape intially printed with 0/blanks. The robot has looked at the symbol in the window (symbol 0), has read the instruction ("state") C and is about to PRINT a 1. Then it will push the tape-LEFT button. Lastly it will look toward instruction ("state") B. (The print/erase mechanism is out of sight, beneath the window.)
"Busy beavers" is a game. The goal? To find "the instructions" that cause you as the busy beaver to print the most ones on a piece of paper tape. But like all games there are rules, and to play, first you will need to learn them.
Rule #1: Know what a "two-symbol Turing machine" is. The Turing machine is NOT easy to understand. But think of it as a really weird, clunky, simple-minded calculating-machine, the most difficult-to-use pocket calculator on the planet, but the most powerful. It has a hugely-long piece of paper tape (marked off into "squares") running through it, and 4 buttons: PRINT, ERASE, LEFT ONE SQUARE, RIGHT ONE SQUARE,  a robot that pushes the buttons, and a list of instructions called "the Table" that the robot must follow. 
As a busy beaver you will be the robot. You have the tape running past you. You will have a list of instructions. You (as a two-state busy-beaver) can PRINT only 1's (tally marks) or ERASE them (in other words, make blanks, print 0's) on this all-blank (all-0's) tape. You can print or overprint, erase or overerase, but only one mark on one square at a time. You can move the tape only one square LEFT or RIGHT at a time. The tape is as long as you want it to be. If you would rather, you can use the push-button console rather than doing this by hand. 
RULE #2: The busy beaver always follows its unchanging list of instructions. You will need to know how to read them and you must follow them without fail. See below, with an example. 
RULE #3: Always obey rules #1 and #2. You are now a computer/robot, after all.
  
If you can do RULE #1, #2 and #3 without mistakes you too can leave the world of humans and join the ranks of the busy beavers! 
Your mission:
Your mission (and your life) as a busy beaver will come in three parts:
Mission Part I:
:> Your mission as a "busy beaver" (should you choose to accept it):
:: >> At the start you will be given a list of instructions. (Thereafter either you or your handlers will change the instructions in Part II). Follow the instructions precisely, and print as many 1's as your instructions tell you to do before halting. To succeed you must print some ones and HALT, eventually!
:> Your "Turing tape" is ready -- it is blank. Your instruction is #1. There is no time limit, no one cares (too much) how long you take. Just follow the rules and print the 1's on your blank tape.
:> Robots, are you ready?
:> Go!
Mission Part II:
Robots, your busy beaver trial is over. You have succeeded: you either came to HALT or you failed to HALT (how did we know this? -- ahh, that's an interesting question!), OR, you printed some ones. The score-keepers are standing by, ready to count your ones and record the instructions that you followed.
:> Your new mission (should you chose to accept it):
:: Change the instructions you were given, so they are still a true "busy beaver Turing program" and they have the same number of instructions. For example: if your mission is to find the best 6 instructions, you must have 6 instructions. But if in instruction #3 you see ERASE, you might change this to PRINT and where you might see LEFT you might change this to RIGHT).
:: do part I again.
Mission Part III:
Repeat Part I and Part II forever! This is the life of a "busy beaver". Aren't you glad you're a robot and not a human!
>>>>>> <<<<<<<<
How to read busy beaver instructions:
Here is the instruction table for a 2-state Turing-machine busy beaver.
                 	Current instruction A:	          	           	Current instruction B:	          	
                 	PRINT/ERASE:          	Move tape:	Next state:	PRINT/ERASE:          	Move tape:	Next state:
tape symbol is 0:	PRINT                 	RIGHT     	B          	PRINT                 	LEFT      	A
tape symbol is 1:	PRINT                 	LeFT      	B          	PRINT                 	NONE      	H
Busy beavers always start at "instruction A" with a blank tape (instructions are usually called "states" in Turing-world). The "HEAD" is where the scanned square is -- where your eye is looking (so "eye" might be a better word).
				HEAD						At start of Instruction:
				    						A
In the instruction table, look down the column on the left. Is the scanned square (in the button console, or before you on the tape, beneath HEAD) blank (or 0), or does it have a 1 written there? It should be blank/0, because we're starting from scratch. Since it is blank/0, under A follow the top row from left to right and do the following in this order: 
: PRINT (mark the square with a 1)
: RIGHT (move the tape to the right)
: GO TO INSTRUCTION B
				HEAD	 					At start of Instruction:
				    	 					A
				    	1					B
At instruction B we look to see if the scanned symbol is a 1 or a blank/0. Since we know it is a blank/0, we follow the top row again, but under B, and do the following:
: PRINT (mark the square)
: LEFT (tape to the left)
: GO TO INSTRUCTION A
			 	HEAD	 					At start of Instruction:
			 	    	 					A
			 	    	1					B
			1	1   	 					A
Now that we find that there's a 1 printed on the scanned square. So we follow the bottom row under A and find the instructions are:
: PRINT
: LEFT
: GO TO INSTRUCTION B
		 	 	HEAD	 					At start of Instruction:
		 	 	    	 					A
		 	 	    	1					B
		 	1	1   	 					A
		1	1	    	 					B
We continue this, when finally we hit the HALT instruction. We're done!:
	 	 	 	HEAD	 					At start of Instruction:
	 	 	 	    	 					A
	 	 	 	    	1					B
	 	 	1	1   	 					A
	 	1	1	    	 					B
	1	1	1	    	 					A
	 	1	1	1   	1					B
	 	1	1	1   	1					H
You as a busy beaver have printed 4 ones. They didn't have to be all in a row, but indeed this is nice work. You have done as well as any "2-state 2-symbol busy beaver" can do. Now its time to go to a three state instruction. Robots: are you ready?
::The end.
>>>>>> <<<<<<<<
An example of a "little" busy beaver --  a one, two, three, or four-state busy beaver -- can be "run" on any spreadsheet such as Excel. Five and six-state busy beavers cannot -- their "productions" -- the numbers of ones they can print -- are too huge to fit. You will need some help setting a model up. It will help you to know what the INDEX(....) instruction does (for example, INDEX(B5:B20,,A3)), but you can make a Turing machine on a spreadsheet without this. Still it's kind of tricky. For an example of what a busy beaver's "run" looks like, see Turing machine examples and Post-Turing machine.wvbaileyWvbailey 17:39, 8 September 2006 (UTC)
thread_title:  The game of Busy Beavers: a simple explanation 

message: thumbnail|900px|Run on Excel. Transposed sideways to fit on page. A. Brady's machine.
The following is a test-case to see what it looks like. It will look better with Netscape viewer. You can find the 2-state 2-symbol busy beaver at Post-Turing machine and as mentioned in the article, the 3-state 2-symbol busy beaver at Turing machine examples. wvbaileyWvbailey 20:37, 22 August 2006 (UTC)
thread_title:  4-state busy beaver example 

message: A few readers (see above) have requested that the article contain some context, explanation and example. I don't feel competent to address the more technical parts here (I could work on the historical -- I want to see Rado's paper, for example). I would suggest the following:
*Make it clear that busy beavers is "a game" (albeit a weird mathematical game) that anyone can play. It has no apparent "use" -- see next bullet-point. 
*Historical context: Why was busy beavers proposed by Rado? Brady hints that it has to do with questions around "the halting problem", instances of tiny machines should be amenable to solution of "the problem" but even these tiny ones are virtually "intractable".
*Who works on busy beavers? The only name I know is Allen H. Brady; cf his paper referenced. 
*A brief description of the algorithm(s) used to find the busiest beaver.
:*How do we know that a particular busy beaver under test (e.g. a 6-state one) is not locked in a loop? How do we know when to "give up" and "call it a day"?
:*Heuristics: are they used? Can a casual reader figure out which b-b's won't work, and why? (certainly some are trivial, such as any accessible state with both 0 and 1 reverting back to itself (to make a circle)
:*Genetic algorithms used in any way? Parallel algorithms used to sift through the possibles? 
*Examples of the simplest cases. For example Post-Turing machine and Turing machine examples where I've used 2- and 3-state two-symbol busy beavers as examples (thus avoiding "original research"). But really those should be repeated here, or especially the busy beaver on Turing machine examples should actually be here, and the Turing machine article refer here.
*Why does the problem "explode?" It seems to explode in two ways: both in the number of possible b-b's per number of states, and (ii) the number of states a particular instance under test might go through.
:*Provide a simple explanation of the number of possible busy beaver machines per number of states N (it looks to me like it is (8*N)^N). But many are silly, hence the need for heuristics.
*Better (more) print references, in-line references too. Any books out there specifically on busy beavers?
Suggestions? Comments? wvbaileyWvbailey 14:47, 15 September 2006 (UTC)
thread_title:  Too technical 

message: I've added quite a bit of new information based upon Rado's original paper and some of your comments on this talk page.  I hope that this has made the page more readable and accessible.  Because of the major content change, I have removed the Confusing tag on this page.  I hope that you all will review the page to see if it ought to be put back or not.  If you do have any comments or requests, I would be very interested in hearing them.
Happy Busy Beavering! Sligocki 17:30, 12 December 2006 (UTC)
replies: text: I very much appreciate what you've done here. I scoured the Dartmouth library for a a cc of Rado's original paper but was unable to find one. I do have a second paper -- the Lin-Rado paper -- but not the very original one. I see someone has added yet another tag at the bottom. But without the original Rado paper I feel unprepared to do an effective review/edit. Where does one find a cc of the original paper? Lemme know, thanks. wvbaileyWvbailey 21:39, 27 January 2007 (UTC)
thread_title:  Content Update 

message: I was wondering if it would be possible to extend the notion of Turing Machines to transfinite numbers of states or symbols.  If, for example, we labeled the states by the natural numbers (instead of alphabetically), and used two algorithms to determine the instructions at each state for symbol 0 and for symbol 1, would it be possible to create a machine that halts after an infinite number of steps and/or prints an infinite number of 1's?  Alternatively, could we use an infinite number of symbols and a finite number of states?  Or a infinite number of symbols and of states?
I haven't thought too long about it, but it seems like it should be possible, perhaps even easy, to create such machines that halt either in a finite or infinite number of steps, or which do not halt.  Ones that do not halt are trivial, and ones that halt in a finite number of steps obviously do not need an infinite number of states or symbols, but ones that halt after some infinite time span might be interesting.
Finally, if such a thing is possible, would it make a difference which infinite number of steps it takes (\omega vs. \omega + 1 vs. \omega^\omega vs. \omega_1, etc.)?  Eebster the Great (talk) 03:49, 1 October 2008 (UTC)
replies: text: This is a very cool idea, but I don't think you'll be able to construct such a thing. Here's the crux of it, what does it mean for a TM to halt after an infinite number of steps? We can easily define the state it will be in at any finite time, but how we define where it will be at time \omega? Perhaps if the TM is in a very simple loop whereby it remains in the same state for all time ≥n , then we could say that it will be in that state at time \omega. But then it won't halt anyway. You'll run into other issues as well, for example, you could run out of tape (it is only \omega long!). Sorry, Sligocki (talk) 12:53, 23 November 2008 (UTC)
replies: text: The idea is that the TM has an infinite number of states, too.  For example, if the TM has \omega states, and each state has instructions following some sort of algorithm that can be generalized to an infinite number of steps and states, then it appears it could run for an infinite number of steps, going through an infinite number of states (perhaps some more than once), and still halt.  The length of tape would not be a problem, because that can always be expanded.  For example, instead of a tape divided into \omega squares, you could have a plane divided into \omega \times \omega squares, and you would be fine.  I'm not saying this generalization is possible (or useful), but it does seem like it ought to be.  Eebster the Great (talk) 21:41, 25 November 2008 (UTC)
replies: text:  You can't exactly do this since a turing machine with infinitely many states can potentially recognize any finite string or go arbitrarily far before stopping. As I understand it you can ask about equivalents of the Busy Beaver when you have access to some reasonably simple Oracle machine. JoshuaZ (talk) 22:00, 25 November 2008 (UTC)
thread_title:  Infinite Busy Beavers? 

message: "There is an analog to the Σ function for Minsky machines ... This is a consequence of the Church-Turing thesis."
Please excuse me if I misunderstand the concepts here.  I thought that the equivalence of register machines and turing machines was a mathematical theorem, whereas the CT-thesis was an unproven philosophical assertion.  
Should it say instead "There is an analog for register machines because somebody proved they are equivalent", perhaps followed by "it is probably impossible to calculate the function by any means because of the CT-thesis."  —Preceding unsigned comment added by 24.2.48.202 (talk) 20:02, 9 January 2009 (UTC)
replies: text: This sounds like a legitimate concern to me. The Church-Turing thesis is not a theorem and we should avoid language referring to it as such. Dcoetzee 21:05, 9 January 2009 (UTC)

text: Yes, I agree and that statement was redundant based on the opening paragraph to the section, so I've removed it. Perhaps something should be said about why there is an analogy to the busy beaver in any formal model of computation. In fact it looks like the Wikipedia article model of computation is rather lacking. Sligocki (talk) 21:03, 15 January 2009 (UTC)
thread_title:  CT-Thesis 

message: The article says: "there is, for each input n, an algorithm An that outputs the number Σ(n) (see examples).". But what's so remarkable about this? This is true of all integer-valued functions, and to define such an algorithm is trivial. - Gigasoft 84.211.109.82 (talk) 04:51, 3 April 2010 (UTC)
replies: text: Imho, the quoted statement is a triviality that deserves to be emphasised for the sake of readers unfamiliar with the subject.  I believe there is a tendency for such readers, upon first encountering the fact that "Σ is not computable", to mistakenly think this implies that there exists some n for which Σ(n) is not computable. — r.e.s. (talk)  13:47, 3 April 2010 (UTC)
replies: text: Ditto. This is a rather subtle distinction and deserves proper emphasis. But feel free to edit the wording if you don't think that the reason for this emphasis is clear. Cheers, — sligocki (talk) 04:29, 8 April 2010 (UTC)
thread_title:  An uninteresting statement 

message: Here was written:
A trivial but noteworthy fact is that every finite sequence of Σ values, such as Σ(0), Σ(1), Σ(2), ..., Σ(n) for any given n, is computable
But why is it "trivial"? Perhaps for some n there would be such a Turing machine that we couldn't identify whether it halts or not.
Eugepros (talk) 11:18, 5 August 2010 (UTC)
replies: text: The noteworthy fact is simply that for any finite n, the sequence Σ(0), Σ(1), Σ(2), ..., Σ(n) is trivially computed by a program such as "PRINT <Σ(0)>, <Σ(1)>, <Σ(2)>, ..., <Σ(n)>", where <x> stands for the decimal representation of x.  E.g., as mentioned in Computable_function#Examples, "PRINT 0, 1, 4, 6, 13" trivially computes Σ(0), Σ(1), Σ(2), Σ(3), Σ(4).  A similar program exists for every finite n, whereas no such program exists for the entire infinite sequence (because a program is by definition a finite-length string).  Perhaps this fact should not be called trivial, as it's the computation involved that's trivial. I've reworded the sentence accordingly. — r.e.s. (talk)  16:22, 9 August 2010 (UTC)

text: Another point ... A BB Turing machine that leaves Σ(n) '1's on the tape might not serve as the very algorithm for Σ(n), because standard conventions (unlike the BB game) require the '1's that encode output to be in a contiguous block. (edited) — r.e.s. (talk)  01:51, 10 August 2010 (UTC)

text: I think that when you say "And we can face fundamental mathematical problems, trying to find the number Σ(10), because halting problem for some Turing machine (of 10 states) might be unresolvable. Can't we?", you are actually asking the very kind of question that concerned Rádo & Lin in the first place.  As I noted in the above section Non-computability of Σ: emphasizing a possible pitfall, Rádo & Lin explicitly stated that they were looking for "clues regarding the formulation of a fruitful concept for the effective calculability (and noncalculability) of individual well-defined integers".  As far as I know, no such concept has yet been formulated, and I don't know to what extent constructive logic might play a role. — r.e.s. (talk)  16:49, 10 August 2010 (UTC)

text: Whatever role constructive logic might (or might not) have in this issue, evidently Rádo & Lin themselves regarded it as inadequate to their purposes.  I say this because constructive logic was already well-developed when they wrote (in 1963) that "at present there is no formal concept available for the “effective calculability” of individual well-defined integers like Σ(4),Σ(5), ...", and that therefore "it is of course not possible to state in precise form the conjecture that there exist values of n for which Σ(n) is not effectively calculable."  (Please excuse the bolding, but I think this conjecture is important, and, however ill-formulated, probably deserves to be mentioned in the article.) Also, Rádo had written earlier in the 1962 paper that "this [principle of the largest element] ... may take us well beyond the realm of constructive mathematics", so it seems they were deliberately looking beyond constructive mathematics to formulate the as-yet unavailable concepts.  — r.e.s. (talk)  15:51, 11 August 2010 (UTC)

text: I would say it is definitely wrong, for reasons already given. Rádo & Lin clearly explain that the conjecture could not as-yet be precisely stated, because it involves a concept that could not as-yet be properly formalized.  In particular, for a given value of n, their concept "Σ(n) is not effectively calculable" is evidently not formalized merely by (\not\exists x~x=\Sigma(n)).  Note that constructive logic, the BHK interpretation, and Kleene's realizability interpretation were all well-developed at the time, and it was Kleene who provided Rádo & Lin with the information that their concept could not as-yet be properly formulated.  — r.e.s. (talk)  17:49, 12 August 2010 (UTC)
thread_title:  Section "Non-computability of Σ" 

message: I cannot find any mention of the initial state
(the start state) with which "running the machine" has to begin with.  Or am I just blind?
— H.Marxen (talk) 16:35, 26 August 2010 (UTC)
replies: text: It seems to have been an omission. I've made it explicit, and have also given a shot at improving some of the wording in this section ("The busy beaver game").— r.e.s. (talk)  17:48, 27 August 2010 (UTC)
replies: text: Nice rewording. — sligocki (talk) 14:21, 28 August 2010 (UTC)
thread_title:  Start state 

message: Hello
I've seen recently that an edit stating that Σ(2,3) = 9 and S(2,3) = 38 has been reversed since there is "no published proof". However, in Pascal Michel's website there is a reference to Lafitte and Papazian 2007 as a proof accessible at http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.104.3021&rep=rep1&type=pdf#page=231 Is this considered as unconvincing or not published ? Thanks - Yves  —Preceding unsigned comment added by 193.49.219.185 (talk) 09:35, 3 September 2010 (UTC)
replies: text: Thanks for the reference! Yes, I take it as a published proof, and I have noted User:Sligocki, who did the revert. May be he now wants to undo the revert. --H.Marxen (talk) 17:20, 3 September 2010 (UTC)

text: Wait a minute... I have been too fast: the case is not yet completely settled, IMHO. I'm going to follow up with an explanation. --H.Marxen (talk) 12:09, 4 September 2010 (UTC)

text: Hi everybody, sorry to be so late to the game. Like Heiner, I had not heard about Lafitte and Papazian's paper before and only heard about claimed unpublished proofs that the class of 2-state 3-symbol machines was computable. In fact, my father and I have categorized all tree-normal-form 2x3 TMs as well, although we would not be confident enough about them to publish a paper. I'll give a look over the paper, in the mean time, if Heiner or Pascal think this proof holds water, I feel free to add it back. Likewise, I completely agree that we could add a note saying that a proof had been submitted, although not verified. Happy busy beavering, — sligocki (talk) 17:24, 6 September 2010 (UTC)

text: Hello (I'm the same as 193.49.219.185), thank you Shawn for your intervention. Since I'm not specialist in computer science (my domain is materials science) I don't know exactly how peer review proceeds for conference papers such as L&P's. However it can be said that: (i) L&P presented their proof publicly, without being refuted, (ii) the proceedings paper has been reviewed and is thus likely free of gross methodological errors, (iii) the paper hasn't been refuted since then (2007), and (iv) it has been cited by P. Michel as a proof; P. Michel very likely read it thoroughly without finding major objections. Thus, the proof holds water to this extent. But, just as another domain I'm interested in (discovery of superheavy nuclei), though valuable, this proof is still a proposition awaiting independent confirmation. Because of its inherent complexity, of the scarcity of specialists and the lack of specialist time, AFAIK confirmation could take a few years. A good example is given in D.Briggs' forum on finishing the (5,2) case: the resolution of the 43 holdouts is just the first step of a proof, necessary for reducing the uncertainty level (as of today, one or more of these holdouts could still finally stop after billions or trillions of steps, exploding the present Σ(5,2)/S(5,2) values) but not sufficient for firmly establishing a proof which complexity level is surely much higher than for the (2,3) case. (btw I would suggest to add D.Briggs' forum as a link). - Yves  —Preceding unsigned comment added by 77.196.150.149 (talk) 21:24, 9 September 2010 (UTC)
replies: text: I still have not had a chance to read the paper (always busy), but it does sound like it has gone through some reasonable review. And as I mentioned, I can confirm the facts (up to a small possible error of only enumerating tree-normal form machines). I've reverted my earlier edit to make these exact values until anyone voices objection. Now I must congratulate Lafitte and Papazian :) Cheers, — sligocki (talk) 05:03, 10 September 2010 (UTC)

text: Thanks again. As soon as I have time I intend to (i) add the L&P reference and the caveat, and (ii) add a link to the Briggs' forum for the (5,2) case - Yves  —Preceding unsigned comment added by 193.49.219.185 (talk) 09:07, 10 September 2010 (UTC)

text: Additions done. Hoping the formulation about the caveat isn't heavy or misleading - Yves  —Preceding unsigned comment added by 77.197.251.244 (talk) 13:48, 26 September 2010 (UTC)
thread_title:  Σ(2,3) and S(2,3) 

message: I've just added this section, but the wording is somewhat tricky.  Presently, the conclusion ("in the context of ordinary mathematics, no specific number can be proved to be greater than Σ(10↑↑10)") is somewhat misleading, as it seems to imply that for n = Σ(10↑↑10), not even "n+1 > n" is "provable in ordinary mathematics".  But the intended meaning is that in the given formal system, there is a formula φ(<n>) (where <n> is a unary notation for the number n) that expresses "n > Σ(10↑↑10)", and the conclusion is that no sentence of the form φ(<n>) is provable in that system. — r.e.s. (talk)  19:06, 16 November 2010 (UTC)
thread_title:  [[Busy_beaver#.CE.A3.2C_complexity_and_unprovability|Σ, complexity and unprovability]] 

message: It says this:
"A noteworthy fact is that, theoretically, every finite sequence of Σ values, such as Σ(0), Σ(1), Σ(2), ..., Σ(n) for any given n, is (trivially) computable, even though the infinite sequence Σ is not computable (see computable function examples)."
In my most charitable interpretation of this sentence, it's saying something that's trivially true but perhaps misleading and not too related to the notion of the computability of the actual sequence of Σ. It's kind of like saying that "every digit in the decimal expansion of an uncomputable number is in isolation a computable number, because a 'digit' is a value from 0-9 and hence a natural number and the naturals are computable." This is true, but it's like this random factoid that can easily send someone in the wrong direction. It's misleading because it seems very much like it's saying that one can figure out, even aside from "practicality" restrictions, what the value of Σ(n) is for any n. This is false.
Using our decimal expansion/digit metaphor, it would be like saying that one can figure out WHICH POSITION in the expansion contains which of these computable entities we've called "digits," and that's just not true. If it were true, the number wouldn't be uncomputable at all. Aside from that, the fact that the concept of a "digit" itself, being a value from 0-9, is a computable number is basically irrelevant to what's being discussed here. And in the case of Σ, it's even more irrelevant; if every value of Σ(n) is a finite natural number, then we already know that values of Σ(n) are in the set of all computable numbers by definition. We already went over this when we defined Σ as a function mapping from N -> N. So there's no need to state that because the codomain of Σ is the set of natural numbers, it's also a subset of the set of computable numbers.
But then the next sentence after what's written above is this, which makes it seem like it's saying something that's just plain false (first sentence included again for reference:
"A noteworthy fact is that, theoretically, every finite sequence of Σ values, such as Σ(0), Σ(1), Σ(2), ..., Σ(n) for any given n, is (trivially) computable, even though the infinite sequence Σ is not computable (see computable function examples). Furthermore, for sufficiently small n, it is also practical to compute Σ(n)."
It now seems like it's saying that the problem with computing Σ(n) is that as n increases, the computation involved becomes so complex that it's not really feasible to figure out Σ(100) or something. But that's not really it. It's just not true that we can figure out what Σ(n) is for every choice of n -at all-, independent of any notion of "practicality," on a completely theoretical level, even if we had an infinite amount of time. It's only for some n that we can compute it at all, period.
I see there was some discussion on this before, but it doesn't look like it was resolved. I'm not sure how to change it yet though. I'll post this for now and maybe we can get a discussion on it. I'd like to remove these few sentences because it either seems misleading, unrelated, or false, depending on how I interpret this.
71.230.120.206 (talk) 09:09, 19 January 2012 (UTC)
replies: text: I agree, there was too much focus on how the integers were in-fact computable. I've removed another paragraph that stressed this. How does it look now? Cheers, — sligocki (talk) 23:58, 12 February 2012 (UTC)
thread_title:  Statement about computability of subsets of Σ is messed up and misleading 

message: Even if we ran a Turing machine until the Heat death of the universe, we will get nowhere near when it halts.
Bubby33 (talk) 21:06, 18 April 2015 (UTC)
thread_title:  How do we know the value of BB(6) 

message: The untitled comments below were taken from the top of the page by User:Negrulio on 28-12-2015 --Negrulio (talk) 16:03, 28 December 2015 (UTC)
replies: text: No, there is a 22 state UTM, but it requires the input to be specified on the tape. Busy beaver is run with a blank tape. With no input. In fact, if you look at the definition of the 22 state UTM, and try running it on a blank tape, the behavior will probably be pretty simple because a blank tape probably doesn't encode a very interesting TM input. Please stop editing this page unless you can cite a reliable source. Cheers, — sligocki (talk) 06:36, 17 October 2010 (UTC)

text: Nope. Put simply, if we knew how big a Turing machine we needed to calculate S(n), then we'd be able to calculate S(n). --Ihope127 16:59, 2 April 2006 (UTC)
thread_title:  Further discussion 

message: This edit removed a "non-reliable" source from the article: https://en.wikipedia.org/w/index.php?title=Busy_beaver&diff=756743646
I don't think this is justified. I'm not saying that Wikia is a reliable source. But, what I am saying is that to dismiss this based on the reliability of the source is a genetic fallacy, because the validity of a deductive argument is independent of the reliability of the source of the argument. In other words, a deductive proof is a proof, regardless of where a proof comes from or is published. However, at the same time I do understand that Wikia isn't a WP:RS, and accordingly shouldn't be included in Wikipedia. But, since the removal of this source reduces the quality of the article by adding an uncited non-trivial claim, and it removes a place where people can learn more about what is being said, I think this calls for WP:IAR. IWillBuildTheRoads (talk) 20:07, 8 January 2017 (UTC)
thread_title:  Removal of "Non-reliable" Sources 

message: What is the name of the function that counts the maximum right shift, R(n)? How does it grow proportionally? 91.66.15.241 (talk) 21:08, 13 March 2017 (UTC)
thread_title:  Maximum Tape Length 

message: The article mentions that "a statement of the exact number of steps it takes to reach the Halt state" is necessary because without it "the problem of verifying every potential entry is undecidable", citing the halting problem. However, doesn't the halting problem is that of determining halting of an arbitrary machine on ARBITRARY INPUT, and the latter requirement is key to the proof? 
If there is a generalization of the halting theorem that prohibits creating an algorithm that for one specific input would predict whether an arbitrary machine would stop, such a generalization should definitely be mentioned on the corresponding "halting problem" page, and the current article should link specifically to that generalization. Otherwise the current article needs to reflect that the "exact number of steps" is only required because we don't know an algorithm to check, and do not even know WHETHER such an algorithm could exist.  — Preceding unsigned comment added by 104.53.222.39 (talk) 01:37, 2 June 2020 (UTC)
thread_title:  Requirement of exact number of steps 

message: #include <stdio.h>
#define STATE_N (4)
#define SYMBOL_N (2)
#define TAPE_N (16)
int move_tb[SYMBOL_N][STATE_N] =
    {
        { +1, -1, +1, +1 },
        { -1, -1, -1, +1 }
    };
int state_tb[SYMBOL_N][STATE_N] =
    {
        { 1, 0, -1, 3 },
        { 1, 2,  3, 0 }
    };
int write_tb[SYMBOL_N][STATE_N] =
    {
        { 1, 1, 1, 1 },
        { 1, 0, 1, 0 }
    };
int tape[TAPE_N];
int main(void)
{
    int header = 11;
    int state = 0;
    int step = 0;
    while (1)
    {
        printf("%4d  ", step);
        for (int i = 0; i < TAPE_N; ++i)
            printf("%c%c  ", i == header ? "hABCD"[1 + state] : ' ', "_1"[tape[i]]);
        printf("\n");
        if (state == -1)
            break;
        int read = tape[header];
        tape[header] = write_tb[read][state];
        header += move_tb[read][state];
        state = state_tb[read][state];
        if (header == -1 || header == TAPE_N)
        {
            printf("ERROR: out of tape (header=%d)\n", header);
            break;
        }
        step++;
    }
    return 0;
}
--MkMkMod (talk) 07:53, 17 July 2020 (UTC)
thread_title:  C code (4-state, 2-symbol busy beaver) 

message: I suspect, that S(n) was formerly noncomputable, but in 2013 when the first-order set theory (which is Rayo(n)) was defined, S(n) became computable.
In this video, Carbrickscity said the reason, why Rayo's number is noncomputable. The reason was: "No one will ever define a number with a googol symbols.". The problem seems to be the number of symbols. There is not enough space in the observable universe for a googol symbols. Emk has shown, that Rayo(7901) > S(265536−1), where S(n) is the maximum shifts function. Only 7901 symbols (which can be easily written down) and already 265536−1 states. This is why I suspect, that S(n) became computable, after Rayo(n) was defined.
The same symbol-problem appears on proving TREE(3) in strictly finite mathematics. This would need 2^^1000 symbols. No one can ever finish such a prove, because there is not enough space in the observable universe for this many symbols. This reason was originally written in this video. So, User:84.154.72.51 wrote this idea for the exact value (or at least a really good bound) of TREE(3). The idea is the first-order set theory for TREE(3) in symbols. The first-order set theory is Rayo(n), where n is the number of symbols. Here, it has been shown, that Σ(2000) > Loader's number ≫ TREE(3). Here, it has been shown, that S(n) ≥ Σ(n) for all n. And of course, 265536−1 ≫ 2000. So, the first-order set theory would reduce the "required number of symbols for TREE(3)" from 2^^1000 to less than 7901.  — Preceding unsigned comment added by 80.142.18.145 (talk) 20:41, 25 October 2020 (UTC)
thread_title:  Maximum shifts function S(n) 

message: The 4-state, 2-symbol busy beaver Examples and Visualizations do not agree. The Examples has 1RH for state C symbol 0, and the Visualizations has 0RH for state C symbol 0. The Example states that the 4-state, 2-symbol busy beaver produces 13 ones in 107 steps and it says "see image." The Visualizations only produces 12 ones when it halts. This should be corrected.  — Preceding unsigned comment added by 50.206.176.154 (talk) 04:17, 3 February 2021 (UTC)
replies: text: Could also be a chance to vectorize that diagram! ~~Ebe123~~ → report 17:11, 21 April 2021 (UTC)

text: The visualization uses the state prior to Halt to represent the Halt state. This works for the one-through-three-state diagrams because the state prior to Halt has symbol=1. Unfortunately, this shorthand causes a minor discrepancy in the four-state diagram. I've added a note to the caption to clarify this. 216.243.58.249 (talk) 04:45, 25 September 2022 (UTC)
replies: text: PS: If the Halt state was instead represented by a black circle with no protrusion, the diagrams could be updated to remove this discrepancy. 216.243.58.249 (talk) 04:48, 25 September 2022 (UTC)
thread_title:  Error in Examples 

message: What this article calls Σ(N) (aka "Rado's sigma function") seems to be called BB(N) in many sources: see for example , , , . "BB" seems to be easier to remember than "Σ"; if I'm right that they are the same thing, I propose we change the usage here to replace it. — The Anome (talk) 20:54, 19 October 2023 (UTC)
thread_title:  Proposal: Change of notation 

message: https://www.sligocki.com//2024/05/22/bb-3-4-a14.html
describes a Busy Beaver candidate which puts more than Ackermann(14) symbols on the tape, in fact exactly 
(2 \uparrow ^ {15} 5)+14 non-zero symbols on the tape NadVolum (talk) 11:41, 31 May 2024 (UTC)
thread_title:  3 State 4 symbol Busy Beaver