I just stumbled upon this webpage in which some kid (years ago, presumably in high school) who uses odd slang defined a huge number by putting various operators together and making recursive call after recursive call. He supposes that this is the biggest number anyone’s ever bothered to concisely define. My intuition is that A(g64, g64) (feel free to call it the xkcd number) is bigger. However, intuition is completely useless in this kind of question.

The Clarkkkkson defined:

http://lab6.com/old/school/yearbook/clarkkkkson.html

(And the xkcd number is the result of the Ackermann function with Graham’s Number as both the arguments, as defined in this comic.)

Anyone want to take a crack at setting up some correspondences and demonstrating which is bigger — The Clarkkkkson or the xkcd number?

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These numbers both recursively increase, so it is quite impossible to calculate the values, let alone compare the two. K will always increase, for as long as time exists. A(g64, g64) will always increase, in theory, past time. So, the answer is, the xkcd number will be bigger than the Clarkkkkson, some while after time ends.

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Where A(n) is Ackermann number n, we define L(n) as A(A(A(A(…A(A(n))…) where the number of recursive A() calls is equal to the value n.

The Ludixkcdrous number is defined as L(1043452800), which is the Unix epoch time of the xkcd.com domain registration.

I suspect the Ludixkcdrous number is far larger than any of the previously mentioned numbers.

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Ok. I did a small, simple program to calculate the Ackermann function. This is on C, limited to 2^64-1 because I’m too lazy to search a big integer library.

A(4,1):

[danix@phobos1 ackermann]$ time ./acker 4 1

65533

real 0m54.338s

user 0m48.125s

sys 0m0.237s

A(4,2) seems just to be impossible to compute, or maybe I might be able to, just that I need a big integer lib. I think the process was around 10 MB when I killed it … ow.

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ackermann(4,x) is where ackermann starts getting stupid huge. You’ll not only need a big int class, but a very large amount of space and time to run it.

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for clarification, ackermann(4,2) is (2^65536)-3 or about 2x 10^19729

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I don’t see where he defined the number ‘2’. ‘2’ is technically defined as ‘1+1’ in conventional usage, which he didn’t do himself. So either his number is undefined, or he’s not allowed to use ‘2’.

In the event of the first case, any number is not comparable to his number. In the event of the second, I’d like to submit the following number:

A(A(1+1+1,1+1+1),A(1+1+1,1+1+1))

Which is 32 charachters, and does not use any numbers defined by someone else.

Moral of the story; If the only way you can say you’ve accomplished something is by applying a rule that invalidates what you’ve accomplished, what have you accomplished.

So it’s not really about the really huge number, but the method you use to get the really huge number, and the general idea of ‘the largest number concisely defined’.

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Heh, I once met one of the dudes who made lab6 at a 2600 meeting.

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Hello XKCD. I’m the author of the original Clarkkkkson article. I was a lawless know-it-all 15 year old at the time, and clearly a furious mental masturbator… not that I’ve stopped being any of these things 🙂

I’m enjoying this post and thread immensely, especially Steve’s proof. I didn’t know anything about Graham’s number at the time, so I’m quite pleased that the Clarkkkkson

isactually bigger. And with the ticking time-bomb that is K squaring every day, it’s got a Gmail-style claim to infinity.I can’t say for sure

whyI wrote it, but my best guess is that it was an extension of the lines joke. Me and the other Lab 6 people were falling over ourselves laughing at the idea that Adam had to do so many lines – laughing at the teacher who had undermined his own punishment. Building up a big number on top of the number of lines was as if to say: “Look! You gave him so many lines, you can construct thebiggest number in the worldout of them!”Such passed for logic 🙂

Anyway, I’m a huge fan of XKCD (I did this) but have neglected the blag so far. I promise to try harder in the future!

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Since the Clarkkkson is a constant, it is invalidated because it grows with passage of time; therefore there will be a point where clarkkkson is bigger than the XKCD number.

How about this: 9 -> 9 -> 9 -> 9 -> 9 -> 9 -> 9 -> 9? That’s definitely bigger than either number.

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Oh, sorry, I meant “not a constant” for the clarkkkson.

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Not sure if this breaks Clarckson’s rules of only using what you defined, but how about this?

ck(A(g_64, g_64), A(g_64, g_64), A(g_64, g_64), A(g_64, g_64))

And no pesky Kappa to determine

Or, if you prefer:

A(ck(A(g_64, g_64), A(g_64, g_64), A(g_64, g_64), A(g_64, g_64)), ck(A(g_64, g_64), A(g_64, g_64), A(g_64, g_64), A(g_64, g_64)))

Ackerman’s function using 2 Clarkkkkson functions each using 4 xkcd numbers.

Compute that!

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I like that one Moser, but I think this might beat it.

BB_BB(g_64)(A(xkcd,xkcd))

That would be the Busy Beaver, to the order of the Busy Beaver function of Graham’s number, function of Ackerman’s function given two xkcd numbers.

But on another note; the Clarkkkkson number kind of cheats because it isn’t a solid unchanging number.

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Just a question, for those not quite as gifted: in the search of big numbers, where does, say, the Goedel self-referential phrase-code for PM fall in line with something like Graham’s Number? And then if we incorporate that number as an axiom, then how large (relatively speaking) would the next resulting self-referential code-phrase be?

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Is this number larger than clarkkkson?

Let’s say that 9 pentated to 9, for example is 9 tetrated to 9 9 times , 9 sextated to 9 is 9 pentated to 9 9 times and so the steps go on.

Now, what if we have xkcd steps (name it xkcd-tation) and we xkcd-tate xkcd xkcd times?It is larger than the Clarkkkson?

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Sigh; just read this thread. In investigating *another* number theory problem, had developed a series of “hyperfactorials” just like the Clarkkksons — but last year (about 2007 Feb). I’d suspected they’d been fooled around with already, but now I’ve got a scintilla of evidence that someone else has been pondering the same number set. The author seems to be referring to another, previous person wwho’d dubbed them “hyperfactorial,” so now the hunt is on to find that person…

(I am a math teacher and by way of avocation, an amateur statistician/probabilist)

So, I get what he’s saying perfectly (once he gets beyond the “lynz” set up). Hyperfactorials get really, nicely huge fairly quickly. I think of them as mathematical poetry (no use for them, yet, although I can envision possible use in computer diagnotics).

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The Conway chained arrow notation, mentioned already, is astonishingly powerful. The simple expression 4 -> 4 -> 4 -> 4 is already far larger than Graham’s Number. In fact, it’s larger than the xkcd number itself.

I can’t tell whether or not it’s larger than the Clarkkkkson too…

but once the chain reaches a fifth term I’m certain it would be.

Therefore “Up” proposes a new number, in honour of the xkcd number, called the Upxkcd number:

http://azureworld.blogspot.com/2008/07/up-function.html

P.S. the uncomputables (such as busy beaver, see four comments up) will probably win this game hands down.

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Eh, someone beat me to the punch with using cardinals… I am quite the dilettant when it comes to sophisticated mathematics, and don’t know if this is valid, but I was going to suggest this “cheat”: aleph-null -1.

P.S. Can’t help but feel a bit disappointed that xkcd number lost, but I found it, somehow, satisfying to see people use the Conway chains to reslove it.

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Associating power towers as ((((n^n)^n)^n)^n)… is inferior to right associativity because it fails to truly define a new function. For instance, a power tower of hyper4 (6, 6) (that is to say, 6 tetrated by 6) with the left associativity gives (((((6^6)^6)^6)^6)^6). When you work out some simple power math, that is the same thing as 6^(6^5). So when you get something like 100 tetrated by 100, rather than having a power tower too large to write, you have 100^(100^99).

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I told a few people today about this argument and 4 of them said “infinity!!” to which i replied, “well, that’s not really a number”.. and the response was a disappointing “infinity minus one!”…. Seriously, try casting it first. lol

big_number = (int)#INF;

pwnd.

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Some of the work by Harvey Friedman involves *very* large numbers:

For example, define n(k) as the length of the longest possible sequence x[1],…,x[n] (of k elements) such that for no i L(A(187196)). Now that is a very large number.

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Also check out Jonathan Bowers’ website: http://www.polytope.net/hedrondude/array.htm

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Ugh, my first post got messed up.

For example, define n(k) as the length of the longest possible sequence x[1],…,x[n] (of k elements) such that for no i < j less than or equal to n/2 is x[i],…,x[2i] a subsequence of x[j],…,x[2j].

It is easily seen that n(1) = 3 (AAA) and n(2) = 11 (ABBBAAAAAAA).

However, n(3) is between A(7183,158386) and A(A(5)). A lower bound for n(4), using Ludixkcdrous’ L(n) function, is L(A(187196)).

L(A(187196)). Now that is a very large number.

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the latest comic made me fall a little in love with you.

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Well, The Name I Was Given’s Busy Beaver-based number wins over every other constant mentioned in this thread, and it’s actually easy to show. All the other numbers are computable, and TNIWG’s number is larger than any computable number expressible in this Universe. It doesn’t matter how clever you are with recursion and funky sequence properties and such… anything not involving BB loses. 🙂

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Ok, now; you owe me a new head.

Just popped back in to this thread I blogged when it was relatively new, and got about halfway down the comments, and my old one seems to have exploded.

And I’d just put out the trash, too.

Pants.

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@SnapDragon – Well, not *everything* loses to BB.

Let n be the largest number that is the unique number satisfying

some formula in the first order language of arithmetic with less

than a google characters.

This beats anything you can generate by busy beaver applications…

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hmm, ok, &= one googolplex

then substitute this value into the expression

hypf(hypf(&,&,&),hypf(&,&,&),hypf(&,&,&) hereon defined as @

Then we have

hypf(hypf(@,@,@)hypf(@,@,@)hypf(@,@,@)

This is the base for what is now defined as $

every attosecond this happens to the number $

hypf($,$,$)

Thereby completely outpacing the Clarkkkkson number and growing at a much faster rate than anything ever before.

I’m actually not very good at maths and can anyone tell me if this works? If it does it stumps me why nobody thought of it yet.

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While referencing the Busy Beaver function isn’t exactly cheating, you’re referring to the maximum possible number of steps any describable function of a certain complexity could actually take, not defining a function itself.

It’s the technical way of saying “the biggest describable number + 1”. Literally.

I’ve been playing with a computable function (I call it the up function) I made that can generate some numbers larger than can be represented even by itself with smaller terms combined with any notation I have ever heard of. The nifty thing is it can also represent many NORMAL numbers. For example, up(1,0,3) = 4, up(3,0,3) = 12, up(1,0,m,n) ~ A(m,n) – it’s almost exactly equal, but a tiny bit off on larger m values.

On the other hand, the only way I have come up with to describe the scale of up(3,2,3) is up(3,2,3). The second term of it is a real doozie – I call it the hyperrepeater – such that up(3,3,3) cannot be described in even in relation to up(3,2,3). Not even up(3,2,up(3,2,up(3,2,3))), or any traditionally way of noting how many times you substitute it into the third or first term.

Not even something like this:

up(3,2,up(3,2,up…up(3,2,3)…))

________________/

L^up(3,2,3)(up(3,2,3))

– using the L function mentioned in a previous post here. That’s up(3,2,3) being called recursively into the L function “up(3,2,3) times” as the base for how many times to recursively call up(3,2,3) into it’s third term. Still insignificant next to up(3,3,3).

Calling a larger number into the second term is quite meaningless.

It does not call on any other functions, it is fully computable, and the only operator it uses is addition. If I can ever formalize my notes on it (and make sure I keep all the credit!… and figure out how to describe the numbers it makes… or even understand them.) I may try to publicize it some day.

By the way, L grows somewhat like Grahams numbers. Gn > L(n) > G(n+1)

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Er, Gn < L(n) < G(n+1). Wasn’t paying attention to what I was typing.

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Er, also, I’m not the same Steve that posted earlier.

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Is G–>G–>G–>G–>G–>G–>G–>G–>G

greater than the number of times any conceivably measurable particle in the universe has moved a distance of 1 Plank length since the Big Bang?

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Or guess I should ask first, is Graham’s number by itself already that big?

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Grahams number is much larger than the odds for the entire universe to have ‘randomly’ quantum tunneled to the exact same spot, for every possible quantum state of the entire universe, all at once, every plank time from the big bang until the heat death of the universe.

“Much larger” itself being a laughable understatement. Consider:

G1 = 3^^^^3 = 3^^^(3^^^3)

3^^^3 =

3^3^…^3^3

_________/

3^3^3 //This is the LENGTH of the layer above.

____/

3

Where 3^3^3 is 7.6 trillion, 3^3^3^3 is larger than a googol (which is larger than the number of elemental particles in the universe), 3^3^3^3^3 is larger than a googolplex (more 0s than there are elemental particles in the universe) and each of the 7.6 trillion 3s is another complete and total change in scope. Then we substitute that in:

3^^^(here)

3^^^(that big number) means:

3^3^…^3^3

_________/

3^3^…^3^3

_________/

3^3^…^3^3

_________/

…

… Using the ‘big number’ as the number of layers.

…

3^3^3

____/

3

That’s G1. G2 has G1 arrows.

3^^^^^3 = 3^^^^(3^^^^3)

3^^^^^^3 = 3^^^^^(3^^^^^3)

…

3^^^… G1 times … ^^^3

Each additional arrow makes the previous step trivial by comparison.

G3 has G2 arrows.

Etc.

And yet, given all that, much larger numbers have been devised, where G64 itself is insufficient for describing how much of an understatement “much larger” is. If you wish to see a cleaner visual representation of G1, you can check the wiki entry here:

http://en.wikipedia.org/wiki/Graham's_number

Grahams number is so large because it is an upper bound on the concept of forming a sub graph in one of an arbitrary number of dimensions in a graph so large that even a purely chaotic distribution must have some small sections of order within it.

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That would be the Busy Beaver, to the order of the Busy Beaver function of Graham’s number, function of Ackerman’s function given two xkcd numbers.

But on another note; the Clarkkkkson number kind of cheats because it isn’t a solid unchanging number.

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Bussy beaver numbers are quite real stuff, see: http://en.wikipedia.org/wiki/Busy_beaver

However, there is still no algorithm found to calculate BB_BB(g_64)(A(xkcd,xkcd)).

The highest BB number found after an awful lot of work is BB(4) which equals 13, so bad luck there. BB(5) is still not found, too difficult.

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clark(x)=Clarkkkkson number, but replacing all references to Kappa by x

Raptor(a,b)=raptor(clark(a),b-1) until b=0 (requiring b to be a counting number)

my number=raptor(xkcd number, xkcd number)

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That would be the Busy Beaver, to the order of the Busy Beaver function of Graham’s number, function of Ackerman’s function given two xkcd numbers.

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I not sure I quite follow your post?? would you mind elaborating more?

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Hi, Can I Find out detailed information about the topic.

mantolamaI could not understand. ?LikeLike

There are some really good points you made in your post…very insightful

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hmm…. i’m pretty sure TREE(xkcd) would dwarf any number here, despite it’s ease in notation (go here: http://en.wikipedia.org/wiki/Kruskal%27s_tree_theorem for more on the TREE function)

Seeing as TREE(3) already dwarfs Graham’s number likely more then Graham’s number dwarfs googolplex, I’m pretty sure TREE(xkcd) would be enough to dwarf all the other numbers here, probably including BB(xkcd). not sure about how one would go about proving this though…

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What about numbers related to the Goodstein sequences?

(e.g. J(k) := number of steps it takes for the Goodstein sequence of 100 to go down to 0; then define S(n,k) := J(J(…(J(k))…)) (J repeated n times). BIG NUMBER:

J(100, 100)

)

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udqbpn = largest non infinite number

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==udqbpn[sigma]1/x

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I can state than Clarkkkkson > xkcd number (using Bowers’ array notation):

Consider the Graham’s number: it is between {3,65,1,2} and {3,66,1,2}.

{3,n,1,2} is approximately Ackermann function plugged to 3 (n times), and so xkcd number is merely between {3,66,1,2} and {3,67,1,2}.

Next the Clarkkkkson (I will set the lower bound for the time being, it will be enough that show this comparison):

hypf(n,2,n) > n^n

hypf(n,3,n) > n^^n

hypf(n,n,n) > n{n-1}n

ck(n,n,n,2) = hypf(n,n,hypf(n,n,n)) > (n{n-1}n){n-1}n > n{n-1}(n+1)

ck(n,n,n,3) > n{n-1}(n+2)

ck(n,n,n,n) > n{n-1}(2n-1)

And A-sequence:

A1 = ck(100,100,100,100) > 100{99}199 > 10{10}10 = {10,2,1,2}

A2 = ck(A1,A1,A1,A1) > 10{10{10}10}10 = {10,3,1,2}

A3 = ck(A2,A2,A2,A2) > {10,4,1,2}

…

A100 > {10,101,1,2}

A101 > {10,102,1,2}

It is only initial value for Clarkkkson, but it is already bigger than xkcd number.

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Wait, no. Clarkkkkson was defined, using the weak operators such as: 4vv4 = ((4^4)^4)^4. I reconsider this.

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