## My number is bigger!

For all your silly time-killing forum games.

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### Re: My number is bigger!

Deedlit wrote:It is "infinite" because (1,0) > (0,n) for any natural number n
Okay, I think I finally understand what you were doing. You're not computing Aleph-0 at all, you're taking it as a basic entity, isomorphic to another dimension. Looking back at your previous post "I take it you have no objections to using ordered pairs of natural numbers? Then you should have no objection to using the ordinal omega^2, which is exactly the same thing!" No serious objection, though it seems (to me) sloppy to conflate trees and numbers like that. Though seeing as you're drawing on existing work, I can hardly blame you.

What I mean is that, for any natural number n, we can call the function H_n or I_n and it will be defined. Are you saying that if one is concerned with computation, it is not normal to consider the set of functions H_n defined? Sure there is a cost of O(f(n)) to compute the function - that just confirms that the function is in fact defined for all n.

You have the first pass definitions of all of H_n and I_n (by virtue of all H_n sharing the same first pass definition), but not the second pass definitions. Or to rephrase, you effectively have two functions metaH(n) and metaL(n) that return H_n and L_n. You have the full definitions of metaH and metaL, but you don't have the second pass of H_i until you execute metaH on i, and so on with metaL and l_n. So in addition computing the result of H_n(X), one must compute the function H_n itself, so in the strictest sense you only "have" four functions so far (two meta functions and two base case functions). I'd say for pure math it's perfectly fine to call H_n and I_n "defined" (first past definitions and computable, if not computed, second pass definitions).

C(x,y,z) is a finite set whenever y and z are finite, which they always are when I call m(x,y,z). So we simply take the largest element of a finite set.
That's fine then.

No, it's not begging the question at all. I don't mean sticking in aleph0 into your function; I mean that your function is an ordered function on triplets of natural numbers, which is the same thing as omega^3.
By begging the question, I meant that you were defining transfinites as ordered tuples. When you originally said "...of the form omega*a + b, which we can think of as simply an ordered pair (a,b)" you seemed to be substituting the referent for the symbol; though now it seems (in context) omega represents "(1,0)", not the other way around.

Is the latter H function computable or not?
Yes, but the use of omega can be removed entirely. Which goes back to my original question "...that number is computable? You reference many uncomputable "numbers" and sets." It now seems that your number can be defined without dragging in the concept of infinity, although not preferable to one familiar with the same work as you.

Thank you for your explanations Deedlit.
The thing about recursion problems is that they tend to contain other recursion problems.

orangedragonfire
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### Re: My number is bigger!

Deedlit wrote:You then define f with a second sequence of natural numbers, but your definitions don't quite work. You start with two cases, k =1 and otherwise, then add two more cases even though "otherwise" should take care of everything.

The two first cases were about how to solve the x-bracket, while the rest is about the y-bracket.

Deedlit wrote:You don't define what "y" is.

It's the size of the y-bracket. Not sure why I forgot to mention that...

Deedlit wrote:And, in your third case, you say:

[math]U_{0, 0, ... , 0, y_{l}, y_{l+1}, ...}(f) = G(f\{0, 0, ... , 0, y_{l}-1, y_{l+1}, ...\}_{[n]})[/math]
[math]f\{0, 0, ... , 0, y_{l}, y_{l+1}, ...\}_{[x_i]} = U_{[0, 0, ... , 0, y_{l}, y_{l+1}, ...]}(f\{0, 0, ... , 0, y_{l}-1, y_{l+1}, ...\}_{[n]})[/math]

you convert from f to [imath]f\{0, 0, ... , 0, y_{l}-1, y_{l+1}, ...\}_{[n]}[/imath] twice, which I'm pretty sure is a mistake.

I think that is correct. In each of those steps, I reduce the size of the first non-zero element by 1, so it should terminate.

Edited definition:

(lists are [imath][x_i][/imath] and [imath][y_j][/imath] of sizes x and y. k is the smallest non-zero index i x-list and l in y-list)

for k = 1 and x > 0:
[math]f\{[y_j]\}_{x_1, x_2, x_3, ...} = U(f\{[y_j]\}_{x_1-1, x_2, x_3...})[/math]
for k > 1 and x >0:
[math]f\{[y_j]\}_{0, 0, ..., 0, x_k, x_{k+1}, x_{k+2}, ...} = G(f\{[y_j]\}_{n, n, ..., n, x_k-1, x_{k+1}, x_{k+2}, ...})[/math]

for x = 0, [imath]y_l[/imath] > 1 or y = 1:
[math]U_{0, 0, ... , 0, y_{l}, y_{l+1}, ...}(f) = G(f\{0, 0, ... , 0, y_{l}-1, y_{l+1}, ...\}_{[n]})[/math]
[math]f\{0, 0, ... , 0, y_{l}, y_{l+1}, ...\}_{[x_i]} = U_{[0, 0, ... , 0, y_{l}, y_{l+1}, ...]}(f\{0, 0, ... , 0, y_{l}-1, y_{l+1}, ...\}_{[n]})[/math]
for x = 0, y > 1 and [imath]y_l[/imath] = 1:
[math]U_{0, 0, ... , 0, 1, y_{l+1}, ...}(f) = G(f\{n, n, ... , n, 0, y_{l+1}, ...\}_{[n]})[/math]
[math]f\{0, 0, ... , 0, 1, y_{l+1}, ...\} = U_{[0, 0, ... , 0, y_{l}, y_{l+1}, ...]}(f\{n, n, ... , n, 0, y_{l+1}, ...\}_{[n]})[/math]

Deedlit
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### Re: My number is bigger!

orangedragonfire wrote:
Deedlit wrote:And, in your third case, you say:

[math]U_{0, 0, ... , 0, y_{l}, y_{l+1}, ...}(f) = G(f\{0, 0, ... , 0, y_{l}-1, y_{l+1}, ...\}_{[n]})[/math]
[math]f\{0, 0, ... , 0, y_{l}, y_{l+1}, ...\}_{[x_i]} = U_{[0, 0, ... , 0, y_{l}, y_{l+1}, ...]}(f\{0, 0, ... , 0, y_{l}-1, y_{l+1}, ...\}_{[n]})[/math]

you convert from f to [imath]f\{0, 0, ... , 0, y_{l}-1, y_{l+1}, ...\}_{[n]}[/imath] twice, which I'm pretty sure is a mistake.

I think that is correct. In each of those steps, I reduce the size of the first non-zero element by 1, so it should terminate.

Your definition doesn't work when yl = 1. First you reduce yl to 0, then you reduce it to -1. I'm pretty sure you don't mean to reduce the same element twice.

for x = 0, y > 1 and [imath]y_l[/imath] = 1:
[math]U_{0, 0, ... , 0, 1, y_{l+1}, ...}(f) = G(f\{n, n, ... , n, 0, y_{l+1}, ...\}_{[n]})[/math]
[math]f\{0, 0, ... , 0, 1, y_{l+1}, ...\} = U_{[0, 0, ... , 0, y_{l}, y_{l+1}, ...]}(f\{n, n, ... , n, 0, y_{l+1}, ...\}_{[n]})[/math]

This seems very confused to me. you define [imath]U_{0, 0, ... , 0, 1, y_{l+1}, ...}(f)[/imath] when y > 1 and yl = 1, but then your second line you apply [imath]U_{0, 0, ... , 0, 1, y_{l+1}, ...}[/imath] to f where you hardcode n into the first argument, so yl = n, which is not necessarily 1. I'm pretty sure you're not meaning what you are saying; your abstractions of U and G are throwing you off.

Deedlit
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### Re: My number is bigger!

Deedlit wrote:It is "infinite" because (1,0) > (0,n) for any natural number n
Okay, I think I finally understand what you were doing. You're not computing Aleph-0 at all, you're taking it as a basic entity, isomorphic to another dimension. Looking back at your previous post "I take it you have no objections to using ordered pairs of natural numbers? Then you should have no objection to using the ordinal omega^2, which is exactly the same thing!" No serious objection, though it seems (to me) sloppy to conflate trees and numbers like that. Though seeing as you're drawing on existing work, I can hardly blame you.

Yes, we do not "compute" any of the ordinals in the manner you describe. I'm not sure what you mean by "conflate trees and numbers". If you mean my identification of omega^2 with ordered pairs, my point was that they are equivalent structures, even if you choose to regard them as separate entities. One may as well regard ordered pairs in parentheses (a,b) as different from ordered pairs [a,b] in brackets. Why bother? Especially when the discussion is of whether a structure is somehow illegal because it brings in the notion of infinity; omega*a + b does not bring in infinity any more into play than (a,b), because they are equivalent structures.

What I mean is that, for any natural number n, we can call the function H_n or I_n and it will be defined. Are you saying that if one is concerned with computation, it is not normal to consider the set of functions H_n defined? Sure there is a cost of O(f(n)) to compute the function - that just confirms that the function is in fact defined for all n.

You have the first pass definitions of all of H_n and I_n (by virtue of all H_n sharing the same first pass definition), but not the second pass definitions. Or to rephrase, you effectively have two functions metaH(n) and metaL(n) that return H_n and L_n. You have the full definitions of metaH and metaL, but you don't have the second pass of H_i until you execute metaH on i, and so on with metaL and l_n. So in addition computing the result of H_n(X), one must compute the function H_n itself, so in the strictest sense you only "have" four functions so far (two meta functions and two base case functions). I'd say for pure math it's perfectly fine to call H_n and I_n "defined" (first past definitions and computable, if not computed, second pass definitions).

Eh, all that matters is that for any specific natural numbers n and x, one can compute H_n(x) and I_n(x). You would accept that, right?

No, it's not begging the question at all. I don't mean sticking in aleph0 into your function; I mean that your function is an ordered function on triplets of natural numbers, which is the same thing as omega^3.
By begging the question, I meant that you were defining transfinites as ordered tuples. When you originally said "...of the form omega*a + b, which we can think of as simply an ordered pair (a,b)" you seemed to be substituting the referent for the symbol; though now it seems (in context) omega represents "(1,0)", not the other way around.

I don't really think of it like that. Omega does not require (1,0) for its definition, any more than (1,0) requires omega for its definition. But in the isomorphism between omega^n and ordered n-tuples, omega is identified with (1,0), and both are greater than an infinite number of elements, so in that sense both are equally "infinite".

Is the latter H function computable or not?
Yes, but the use of omega can be removed entirely. Which goes back to my original question "...that number is computable? You reference many uncomputable "numbers" and sets." It now seems that your number can be defined without dragging in the concept of infinity, although not preferable to one familiar with the same work as you.

Yes, any mention of ordinals can be removed from the definition. However, my point is that infinity is nonetheless _implicit_ in structures such as ordered tuples (or even the natural numbers, which has order type omega). If you don't regard a structure having an infinite order type an introduction of infinity, then recursive ordinals do not introduce infinity either. So I could not use the word "ordinal" anywhere, and always use "finite tree structures", but the thing is these "finite tree structures" would be exactly equivalent to ordinals. So it's just a matter of calling a spade a spade. Note that to define the larger ordinals that I need, you need to do things like define normal functions on ordinals, and take fixed points. So we really need a definition equivalent to ordinals; I could, say, define "well-ordered finite tree structures", and then take equivalence classes of order types - but that is exactly what an ordinal is! So avoiding ordinals is really just playing with words.

Thank you for your explanations Deedlit.

No sweat! I think we are starting to converge here.

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### Re: My number is bigger!

Deedlit wrote:Yes, we do not "compute" any of the ordinals in the manner you describe. I'm not sure what you mean by "conflate trees and numbers". If you mean my identification of omega^2 with ordered pairs, my point was that they are equivalent structures, even if you choose to regard them as separate entities. One may as well regard ordered pairs in parentheses (a,b) as different from ordered pairs [a,b] in brackets. Why bother? Especially when the discussion is of whether a structure is somehow illegal because it brings in the notion of infinity; omega*a + b does not bring in infinity any more into play than (a,b), because they are equivalent structures.

"omega^2", "(a,b)", and "[a,b]" are not equivalent, they are isomorphic for the operations you were performing. Being as that we had difficulty establishing common context, the gap between equivalence and isomorphism is important. When I first saw "Limit cardinal" it was defined with aleph-0, which was in turn defined with |natural numbers|, a computationally impossible operation.

Eh, all that matters is that for any specific natural numbers n and x, one can compute H_n(x) and I_n(x). You would accept that, right?
Yes, I will. However I (personally) wouldn't "say" we have an infinite number of functions (in past or present tense). However we've narrowed that down to an issue of semantics.

I don't really think of it like that. Omega does not require (1,0) for its definition, any more than (1,0) requires omega for its definition. But in the isomorphism between omega^n and ordered n-tuples, omega is identified with (1,0), and both are greater than an infinite number of elements, so in that sense both are equally "infinite".

"Omega = |set of 2-tuples of natural numbers < (1,0) under this ordering|" and "Omega^2 = |set of 2-tuples of natural numbers|" seems to be the connections you're making. This connection can be made validly and consistently, but it's not necessary under the definitions of either; it's a subjective choice. One could also have made the connection "Omega = |set of computer programs more inefficient than windows calculator|", therefore "Omega = windows calculator"

Yes, any mention of ordinals can be removed from the definition. However, my point is that infinity is nonetheless _implicit_ in structures such as ordered tuples (or even the natural numbers, which has order type omega).

Is it implicit up (can be derived from) or implicit down (part of the definition or derivation of a tuple)? If up, several problems are raised: First, indicating which implication you have chosen to equal tuples. Second, does going from A to B do anything needed? Third, if a reader assumes the A -> B connection's inclusion means that it is both relevant and strictly necessary, what might they presume?

So we really need a definition equivalent to ordinals; I could, say, define "well-ordered finite tree structures", and then take equivalence classes of order types - but that is exactly what an ordinal is! So avoiding ordinals is really just playing with words.

I agree we need a definition isomorphic to how you're using ordinals. But I don't see it a necessary to describe the order types, then the equivalence class of said order types, when one could start with modded out definitions.

BTW, I'm working out a rephrasing of your number post for 1) making sure I understand you correctly 2) providing you with examples of why rephrasing without reference to infinity would be beneficial. However I'm stuck on the line "?(y, z) = min{a : z ? C(y,a,?), C(y,a,?) n z ? a}", and having trouble looking up that usage of the question mark.
The thing about recursion problems is that they tend to contain other recursion problems.

tastelikecoke
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### Re: My number is bigger!

I'll be ignoring the first few passages with this one:

Let function M(x) be Ackermann function(Ackermann function(Ackermann function(x))).

let D be the number of distinct solutions of the N-queens puzzle in the 9235th dimension, where N is

M(M( M( M( M( M(9325) ) ) ) ) )

Now, let function K(x) be the xth digit of D in decimal.
Let there be D nodes, where every node is connected other nodes such that the edge between Nodea and Nodeb is equal to K(b) - K(a).

Let P be the amount of paths from Node1 to NodeD where the total weight sum of all edges crossed is greater than 0.

let Q is the amount of possible tickets to draw on a P/P lottery (where the you can choose one out of P numbers for P sections of the ticket)

My number is M(M(Q)).

(by the way, I haven't checked what the upper bound of my number would be )

Deedlit
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### Re: My number is bigger!

Quizatzhaderac wrote:"omega^2", "(a,b)", and "[a,b]" are not equivalent, they are isomorphic for the operations you were performing. Being as that we had difficulty establishing common context, the gap between equivalence and isomorphism is important. When I first saw "Limit cardinal" it was defined with aleph-0, which was in turn defined with |natural numbers|, a computationally impossible operation.

I think you are using isomorphic in the same way I am using equivalent.

Yes, I will. However I (personally) wouldn't "say" we have an infinite number of functions (in past or present tense). However we've narrowed that down to an issue of semantics.

Would you say the function H_0(x) has an infinite domain?

I don't really think of it like that. Omega does not require (1,0) for its definition, any more than (1,0) requires omega for its definition. But in the isomorphism between omega^n and ordered n-tuples, omega is identified with (1,0), and both are greater than an infinite number of elements, so in that sense both are equally "infinite".

"Omega = |set of 2-tuples of natural numbers < (1,0) under this ordering|" and "Omega^2 = |set of 2-tuples of natural numbers|" seems to be the connections you're making. This connection can be made validly and consistently, but it's not necessary under the definitions of either; it's a subjective choice. One could also have made the connection "Omega = |set of computer programs more inefficient than windows calculator|", therefore "Omega = windows calculator"

I'm unsure of your argument here. The structures omega^n and ordered n-tuples are isomorphic, so I'm making the "subjective choice" of pointing that out. So if one structure is acceptable, so is the other. It sounds like you are saying "you don't have to make the argument you are making, so it doesn't necessarily hold." Like if I presented a proof, and you replied, "well, you don't have to follow the steps of that proof, its a subjective choice, so it's not really a proof." Or something like that.

Yes, any mention of ordinals can be removed from the definition. However, my point is that infinity is nonetheless _implicit_ in structures such as ordered tuples (or even the natural numbers, which has order type omega).

Is it implicit up (can be derived from) or implicit down (part of the definition or derivation of a tuple)? If up, several problems are raised: First, indicating which implication you have chosen to equal tuples. Second, does going from A to B do anything needed? Third, if a reader assumes the A -> B connection's inclusion means that it is both relevant and strictly necessary, what might they presume?

I don't really understand this either. Implicit up and implicit down?

So we really need a definition equivalent to ordinals; I could, say, define "well-ordered finite tree structures", and then take equivalence classes of order types - but that is exactly what an ordinal is! So avoiding ordinals is really just playing with words.

I agree we need a definition isomorphic to how you're using ordinals. But I don't see it a necessary to describe the order types, then the equivalence class of said order types, when one could start with modded out definitions.

Sure. But, as you say, we need a definition isomorphic to ordinals.

BTW, I'm working out a rephrasing of your number post for 1) making sure I understand you correctly 2) providing you with examples of why rephrasing without reference to infinity would be beneficial. However I'm stuck on the line "?(y, z) = min{a : z ? C(y,a,?), C(y,a,?) n z ? a}", and having trouble looking up that usage of the question mark.

I think maybe your browser is not showing symbols correctly or something - I don't use question marks at all. The line in question is

ψ(y, z) = min{a : z ∈ C(y,a,ω), C(y,a,ω) ∩ z ⊆ a}

If you can't read that, try:

psi(y, z) = min {a : z is an element of C(y, a, omega) and (C(y, a, omega) intersect z) is a subset of a}

Vytron
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### Re: My number is bigger!

Okay, I've read the entire thread, it took a while. I just thought about a uncompressing function that seems huge in my head. It depends on the first numbers that appear in the number, so that all of them uncompress until you're left with a 1 followed with a lot of zeroes.

So, let's define the levels:

Level 1: On this level, if the first digit is higher than 1, you decrease it, and add a number of copies of this number equivalent to itself+1. If the first number is a one, you move to the next one. If the next one is a 0, you move to the next one. If the next one is a 1, you copy the whole sequence but change it so that each number is decreased by 1, unless it's a 0. Example:

232

First digit is >1, so it decreases, and adds 2 ones to the end.

13211

First digit is a 1, so it doesn't decrease. The next one does, it's a 3, so it decreases, and adds 3 2's at the end.

Spoiler:
1221122
112112211
102112211001001100
10111221100100110011
10011221100100110011000001100000000000001001122110010011001100000110000000000000
10001221100100110011000001100000000000000000011000000000000000000000000000000000
...

So, eventually, the 2's disappear, all the copies add 0's, and it halts.

Of note is that the uncompression doesn't depend on the starting value, but on the highest number of its decimal representation. So 9 uncompresses like this:
Spoiler:
9
8888888888
788888888877777777
6888888888777777776666666
5888888888777777776666666555555
...
188888888877777777666666655555544444333322211
17888888887777777766666665555554444433332221177777777
...
11888888887777777766666665555554444433332221177777777...333322211
10888888887777777766666665555554444433332221177777777...33332221100777777776666666655555554444443333322221110066...333322211

Okay, so that's a very puny, let's look at Level 2.

In level 2, you do the same thing, but the number of copies are multiplied by the previous number in the chain, and the number of steps, so:

Spoiler:
Step1:9
Step2:8888888888888888888 (9*2=18 copies of the 8)
Step3:7888888888888888888777...(3*8888888888888888888=26666666666666666664 copies of the 7)...777
Step4:6888888888888888888777...(as step 3)...777666...((The number in step 3)*4 repeating 6's)...666
Step5:5888888888888888888777...(as step 3)...777666...((as step 4)...666555...((The number in step 4)*5 repeating 5's)...55

Level 3: As Level 2. But the previous number in the chain is squared, and the number of steps is exponentiated by that number, and this is the number of copies.

Spoiler:
Step1:9
Step2:888...(92=81;281 8's)...888
Step3:7888...(as step 2)...888777...((3(the number at step 2)^2 7's)...777
Step4:6888...(as step 2)...888777...(as step 3)...777666...((4(the number at step 3)^2 6's)...666

Level 4: As level 3. But the previous number in the chain is exponentiated by itself, and the number of steps builds a tower of exponentations of that length that contain themselves. This is the number of copies.

Spoiler:
Step1:9
Step2:888...(99=387420489;x=387420489;(...((2x)^x)...(x times)...^x) 8's)...888
Step3:788...(as step 2)...888777...(y=S2S2;(...((3y)^y)...(y times)...^y) 7's)...777

Level 5: As level 4, but the previous number in the chain builds a tower of exponentiations of its length that contains itself, and that's the number of up-arrows that the steps use on themselves, and that's the number of copies.

Spoiler:
Step1:9
Step2:888...((...(x=99)^9)^9)^9)^9)^9)^9)^9)^9)); 2↑↑↑...(x times)↑↑↑x 8's)...888
Step3:788...(as step 2)...888777...(y=S2S2)^S2)...(S2 times)...^S2);(...((3↑↑↑...(y times)↑↑↑y 7's)...777

Level 6:

Spoiler:
Step1:9
Step2:888...((...(x=9↑↑↑↑↑↑↑↑↑9); 2(↑↑↑...(x times)↑↑↑x↑↑↑...(x times)↑↑↑x↑↑↑...(x times)↑↑↑x[...x times...]x↑↑↑...(x times)↑↑↑x)x 8's)...888
Step3:788...(as step 2)...888777...(y=S2↑↑↑...(S2 times)...↑↑↑S2); 3(↑↑↑...(y times)↑↑↑y↑↑↑...(y times)↑↑↑y↑↑↑...(y times)↑↑↑y[...y times...]y↑↑↑...(y times)↑↑↑x)y 7's)...777

And so on, and so forth. Now, I'd like to call the xkcdth step of the level xkcd of the xkcdth step of the level xkcd-1 of the xkcdth step of the level xkcd-2...of the xkcdth step of the level 6 of the xkcdth step of the level 5 of the xkcdth step of the level 4 of the xkcdth step of the level 3 of the xkcdth step of the level 2 of the xkcd'th step of level 1, provided all of them are increased until all their digits are 9's at the time that number is called, and this number is added to the number of 9's that a number has, that I use to call in the step xkcdth-1 of the level xkcd of the step xkcdth-1 of level xkcd-1 of the xkcdth-1 step of the level xkcd-2 of the xkcdth-1 step of the level xkcd-3 ... of the xkcdth-1 step of the level 2 of the xkcdth-1 step of level 1, provided all of them are increased until all their digits are 9's at the time that number is called, and this number is added to the number of 9's that a number has, that I use to call in the step xkcdth-2 of the level xkcd of the xkcdth-2 step of the level xkcdth-2 of the level...

...keep talking like this for xkcd more paragraphs...

...provided all of them are increased until all their digits are 9's at the time that number is called, and this number is added to the number of 9's that a number has, that I use to call the first step of the level xkcd of the first step... (all these are 9's, so I only end adding 9 nines at the end of the number, but remember every nine is very important in the explosion) ... of the first step of the level 2 of the first step of level 1, provided all of them are increased until all their digits are 9's at the time that number is called, and this number is added to the number of 9's that a number has, and I call this number A'.

Now, I'd like to call the A'th step of the level A' of the A'th step of the level A'-1 of the A'th step of the level A'-2...of the A'th step of the level 6 of the A'th step of the level 5 of the A'th step of the level 4 of the A'th step of the (this is a copy of paragraph 1, but instead of using xkcd I'm using a much larger number)

...keep talking like this for A' more paragraphs...

... of the first step of the level 2 of the first step of level 1, provided all of them are increased until all their digits are 9's at the time that number is called, and this number is added to the number of 9's that a number has, and I call this number B'.

Now, I'd like to call the B'th step of the level B' of the B'th step of the level B'-1 of the B'th step (this is a copy of paragraph 4, but instead of using A' I'm using a much larger number)

...keep talking like this for B' more paragraphs...

...added to the number of 9's that a number has, and I call this number C'.

Now, I'd like to call the C'th step...
...keep talking like this for C' more paragraphs...
...and I call this number D'.
...keep talking like this for D' more paragraphs...
...keep talking like this for E' more paragraphs...
...keep talking like this for F' more paragraphs...
.
.
.
keep adding symbols, until you have xkcd symbols, let the last symbol be called A''. Now, repeat this whole process (not just the first part) xkcd times, but now, keep adding symbols until you have A' symbols. Now, repeat this whole process until you have 'B symbols. Now, repeat this whole process until you have 'C symbols...

...Now, repeat this whole process until you have A'' symbols, let the last symbol be called B''.
...Now, repeat this whole process until you have B'' symbols, let the last symbol be called C''.
...Now, repeat this whole process until you have C'' symbols, let the last symbol be called D''.
.
.
.

Keep repeating the repetitions, until you have Z'''(Y'''(X'''...A'(xkcd 's)')... 's)... 's)... 's)... 's. Let this symbol be subscript °, so that that last number is equal to A°. Repeat this process until you run out of subscripts (ie. the number of subscripts is A°), and let this be AA which continues like AB ... AA' ... AB' ... AA'' ... A, until AZ°(Y°(X°...A°(xkcd °s)°)... °s)... °s)... °s)... °s, which is equal to BA. Continue like this, until you run out of letters (ie., the number of symbols is AA), and this will be equivalent to A[0]. A[A[0]] = B[0]. Continue like this until you run out of symbols (ie. the number of symbols is A[0]), and the last number will be A<0>. A<A<0>>=B<0>.

...

Continue like this until you run out of symbols to encapsulate the 0 (the number of symbols is some finite amount x so that Z<Z<Z<Z...Y<0> times ...Z<Z<Z<0>>...>> < x < ), and let the highest number of the last symbol be ().

Finally, I'll send my number:

V=‽(‽(...x times]...(‽(‽(999)))))

I just hope this humongous mastodon is at least as big as the numbers people were posting back in page 11...
Go! Go! You can do it username5243!
Cheers Marsh'n!

THANKS KARHELL!!

Deedlit
Posts: 91
Joined: Sun Mar 08, 2009 2:55 am UTC

### Re: My number is bigger!

Vytron wrote:Okay, I've read the entire thread, it took a while. I just thought about a uncompressing function that seems huge in my head. It depends on the first numbers that appear in the number, so that all of them uncompress until you're left with a 1 followed with a lot of zeroes.

So, let's define the levels:

Level 1: On this level, if the first digit is higher than 1, you decrease it, and add a number of copies of this number equivalent to itself+1. If the first number is a one, you move to the next one. If the next one is a 0, you move to the next one. If the next one is a 1, you copy the whole sequence but change it so that each number is decreased by 1, unless it's a 0. Example:

232

First digit is >1, so it decreases, and adds 2 ones to the end.

13211

First digit is a 1, so it doesn't decrease. The next one does, it's a 3, so it decreases, and adds 3 2's at the end.

Spoiler:
1221122
112112211
102112211001001100
10111221100100110011
10011221100100110011000001100000000000001001122110010011001100000110000000000000
10001221100100110011000001100000000000000000011000000000000000000000000000000000
...

So, eventually, the 2's disappear, all the copies add 0's, and it halts.

Okay, let's compute an upper bound to the final number. Let's say you start with n numbers, the largest of which is k. Each number is decreased at most k times, adding at most k numbers to t he sequence, and we do this n times. So we have a maximum of nk2. In addition, we copy the whole sequence at most n times, so we get an upper bound of nk22n. So we have a sequence of length at most nk22n with the largest number k-1. Since n < 2n and k2 <= 2k for k >= 4, we have nk22n <= 22n+k. We know apply the same procedure to the longer sequence; by the same argument, we get a sequence of length at most 22*2[sup]2n+k + k - 1[/sup] < 24*2[sup]2n+k[/sup] = 22[sup]2n+k+2[/sup], of which the largest number is k-2. Applying the same argument again, we get 22*2[sup]2[sup]2n+k+2[/sup]+ k - 3[/sup] < 22[sup]2[sup]2n+k+2 + 2[/sup][/sup] < 22[sup]2[sup]2n+k+3[/sup][/sup], the largest of which is k-3. After k steps, we get a sequence 1000...000 of length less than 2^2^2^...^(2n+2k), with k 2's in the tower.

At later levels you add f(previous number) numbers at each step, where f keeps increasing as you go up levels. Again, starting from n numbers the largest of which is k, we add less than f(10^n) numbers. For the next step, our whole number is less than 10^f(10^n), so we add less than f(10^f(10^n)). Letting F(n) = f(10^n), we have F2(n) = f(10^f(10^n)). After k steps, we get a number at most 10F[sup]k(n)[/sup]. If f(n) is about n↑↑↑...↑↑↑n with m ↑'s, then 10F[sup]k(n)[/sup] will be about n↑↑↑...↑↑↑k with m+1 ↑'s. More generally, if f is at the level of Falpha in the fast-growing hierarchy, then 10F[sup]k(n)[/sup] will be at the level Falpha+1 in the fast-growing hierarchy.

Level 5: As level 4, but the previous number in the chain builds a tower of exponentiations of its length that contains itself, and that's the number of up-arrows that the steps use on themselves, and that's the number of copies.

Spoiler:
Step1:9
Step2:888...((...(x=99)^9)^9)^9)^9)^9)^9)^9)^9)); 2↑↑↑...(x times)↑↑↑x 8's)...888
Step3:788...(as step 2)...888777...(y=S2S2)^S2)...(S2 times)...^S2);(...((3↑↑↑...(y times)↑↑↑y 7's)...777

Level 6:

Spoiler:
Step1:9
Step2:888...((...(x=9↑↑↑↑↑↑↑↑↑9); 2(↑↑↑...(x times)↑↑↑x↑↑↑...(x times)↑↑↑x↑↑↑...(x times)↑↑↑x[...x times...]x↑↑↑...(x times)↑↑↑x)x 8's)...888
Step3:788...(as step 2)...888777...(y=S2↑↑↑...(S2 times)...↑↑↑S2); 3(↑↑↑...(y times)↑↑↑y↑↑↑...(y times)↑↑↑y↑↑↑...(y times)↑↑↑y[...y times...]y↑↑↑...(y times)↑↑↑x)y 7's)...777

And so on, and so forth.

Well, your levels don't really follow a pattern here. The f(n) in levels 2, 3, and 4 correspond more or less to F1, F2, and F3 in the fast-growing hierarchy. Then, at level 5 you jump to step↑↑↑...↑↑↑x with x ↑'s which is equivalent to Fω in the fast-growing hierarchy. In level 6, you take y = previous number↑↑↑...↑↑↑ previous number with previous number ↑'s, then take 3↑↑↑...(y times)↑↑↑y↑↑↑...(y times)↑↑↑y↑↑↑...(y times)↑↑↑y[...y times...]y↑↑↑...(y times)↑↑↑x ~ y↑↑↑...↑↑↑y with y+1 ↑'s. This is less than Fω+1(n), which takes x -> x↑↑↑...↑↑↑x n times, whereas level 6 just takes it twice.

Now, I'd like to call the xkcdth step of the level xkcd of the xkcdth step of the level xkcd-1 of the xkcdth step of the level xkcd-2...of the xkcdth step of the level 6 of the xkcdth step of the level 5 of the xkcdth step of the level 4 of the xkcdth step of the level 3 of the xkcdth step of the level 2 of the xkcd'th step of level 1, provided all of them are increased until all their digits are 9's at the time that number is called, and this number is added to the number of 9's that a number has, that I use to call in the step xkcdth-1 of the level xkcd of the step xkcdth-1 of level xkcd-1 of the xkcdth-1 step of the level xkcd-2 of the xkcdth-1 step of the level xkcd-3 ... of the xkcdth-1 step of the level 2 of the xkcdth-1 step of level 1, provided all of them are increased until all their digits are 9's at the time that number is called, and this number is added to the number of 9's that a number has, that I use to call in the step xkcdth-2 of the level xkcd of the xkcdth-2 step of the level xkcdth-2 of the level...

Not sure what you mean by this. Taking the xkcdth step of level xkcd is not as good as taking the final number of level xkcd - that is the whole point of having a terminating process. I don't know what " xkcdth step of the level xkcd of the xkcdth step of the level xkcd-1" means - the xkcd step of the level xkcd is a number, not a function, so you can't apply it to something else.

...keep talking like this for xkcd more paragraphs...

...provided all of them are increased until all their digits are 9's at the time that number is called, and this number is added to the number of 9's that a number has, that I use to call the first step of the level xkcd of the first step... (all these are 9's, so I only end adding 9 nines at the end of the number, but remember every nine is very important in the explosion) ... of the first step of the level 2 of the first step of level 1, provided all of them are increased until all their digits are 9's at the time that number is called, and this number is added to the number of 9's that a number has, and I call this number A'.

Now, I'd like to call the A'th step of the level A' of the A'th step of the level A'-1 of the A'th step of the level A'-2...of the A'th step of the level 6 of the A'th step of the level 5 of the A'th step of the level 4 of the A'th step of the (this is a copy of paragraph 1, but instead of using xkcd I'm using a much larger number)

...keep talking like this for A' more paragraphs...

... of the first step of the level 2 of the first step of level 1, provided all of them are increased until all their digits are 9's at the time that number is called, and this number is added to the number of 9's that a number has, and I call this number B'.

Now, I'd like to call the B'th step of the level B' of the B'th step of the level B'-1 of the B'th step (this is a copy of paragraph 4, but instead of using A' I'm using a much larger number)

...keep talking like this for B' more paragraphs...

...added to the number of 9's that a number has, and I call this number C'.

Now, I'd like to call the C'th step...
...keep talking like this for C' more paragraphs...
...and I call this number D'.
...keep talking like this for D' more paragraphs...
...keep talking like this for E' more paragraphs...
...keep talking like this for F' more paragraphs...
.
.
.
keep adding symbols, until you have xkcd symbols, let the last symbol be called A''. Now, repeat this whole process (not just the first part) xkcd times, but now, keep adding symbols until you have A' symbols. Now, repeat this whole process until you have 'B symbols. Now, repeat this whole process until you have 'C symbols...

...Now, repeat this whole process until you have A'' symbols, let the last symbol be called B''.
...Now, repeat this whole process until you have B'' symbols, let the last symbol be called C''.
...Now, repeat this whole process until you have C'' symbols, let the last symbol be called D''.
.
.
.

Keep repeating the repetitions, until you have Z'''(Y'''(X'''...A'(xkcd 's)')... 's)... 's)... 's)... 's. Let this symbol be subscript °, so that that last number is equal to A°. Repeat this process until you run out of subscripts (ie. the number of subscripts is A°), and let this be AA which continues like AB ... AA' ... AB' ... AA'' ... A, until AZ°(Y°(X°...A°(xkcd °s)°)... °s)... °s)... °s)... °s, which is equal to BA. Continue like this, until you run out of letters (ie., the number of symbols is AA), and this will be equivalent to A[0]. A[A[0]] = B[0]. Continue like this until you run out of symbols (ie. the number of symbols is A[0]), and the last number will be A<0>. A<A<0>>=B<0>.

...

Continue like this until you run out of symbols to encapsulate the 0 (the number of symbols is some finite amount x so that Z<Z<Z<Z...Y<0> times ...Z<Z<Z<0>>...>> < x < ), and let the highest number of the last symbol be ().

Finally, I'll send my number:

V=‽(‽(...x times]...(‽(‽(999)))))

I just hope this humongous mastodon is at least as big as the numbers people were posting back in page 11...

This could all be made much more clear.

Darvince
Posts: 15
Joined: Wed Oct 10, 2012 1:59 am UTC
Location: Some weird alternate reality that keeps getting worse

### Re: My number is bigger!

A(g64, A(g64, A(g64, A(g64, A(g64, A(g64, A(g64, A(g64, A(g64, g64)))))))))
Last edited by Darvince on Sat Dec 08, 2012 6:14 am UTC, edited 1 time in total.
If it's Good enough for Randall, it's Good enough for Me

(he/him/his)

Vytron
Posts: 429
Joined: Mon Oct 19, 2009 10:11 am UTC
Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P

### Re: My number is bigger!

Okay, thanks for your analysis Deedlit, I think my number is very unclear a this point, I just had to write all that before I forgot about the idea. I think the levels could be made a lot more powerful very easily, but the main power of the number is the recursions that come up later, so I'm not even sure the Levels have to be that strong.

Deedlit wrote:Not sure what you mean by this. Taking the xkcdth step of level xkcd is not as good as taking the final number of level xkcd - that is the whole point of having a terminating process.

Okay, let me clarify. You have Level xkcd, then, you feed some number into it, again, I don't know if the number you feed it matters much, because at some point functions are so strong that they grow to the same size whether you feed them g64 or 2

But just in case, let's define a function. In my above post I said that "provided the number is increased until all of them are nines", which means that if a step returned 1000, this step returns 9999. This is necessary because the levels run on 9's, a 8 on there makes the whole of it much less powerful, but I think I can do better:

9(x)

Returns x times 9. So, 9(9), returns 999999, and 9(1000) returns a string of 1000 9's. Now I define the initial seed as:

۹=9(xkcd)

A string of xkcd 9's. This would be feed into Level 1 instead of just a 9. Just in case this matters at the end.

Now, when I say "take the xkcdth step", I mean, you put ۹ into level xkcd, you take the last number (the uncompressed one, a 1 with a bunch of zeroes) and call it x. You run 9(x), to get a string of 9's that is the length of that number, you call it, huh, ۹[1]. Then you feed ۹[1] into Level xkcd, you take the last number, call it x, run 9(x), to get a string of 9's of that length which is ۹[2]. You feed ۹[2] into Level xkcd. You continue on and on until you get ۹[xkcd]. Let's call it ۶.

Now, for the "xkcdth step of the level xkcd-1", you take ۶ into level xkcd-1, you take the last number x, call 9(x), get ۶[1], feed it into level xkcd-1, and repeat another xkcd times. You get ۶[2]

Now that I (think) I have some clarified recursion, I notice that the levels are getting weaker while the seed is getting bigger. That may not be optimal, maybe the number would be larger if I kept feeding it to level xkcd all the time, but did it one time less.

No, actually, I can keep increasing the levels and decreasing the times, so the number of nines I add at every recursion is greater than the last time, instead of less, and I ensure that it halts.

So, lets redefine "xkcdth step of the level xkcd-1" to be: you take ۶ into level 2xkcd, you take the last number x, call 9(x), get ۶[1], feed it into level 2xkcd, and repeat another xkcd-1 times. You get ۶[2]. Yes, this sounds better. You continue until you get ۶[2xkcd], this will be ۹ (yeah,I don't think people have used colors in the thread to define numbers, I think I'll use that for an extra set of recursive recursions at the end).

Now, for "xkcdth step of the level xkcd-2", use ۹ as the seed, feed it into level xkcd2, call 9(x), get ۹[1], and repeat (by repeat I don't mean you feed ۹ into itself that many times, you have to take ۹[1], for the next repetition, and then ۹[2], for the next repetition) another xkcd-2 times. At the end, you get ۶.

"xkcdth step of the level xkcd-3", is like that, but it's feed to the level xkcdxkcd xkcd-3 times. "xkcdth step of the level xkcd-4" is feed on level (...((xkcd)^xkcd)^xkcd)^...(xkcd times)...^xkcd)^xkcd xkcd-4 times. "xkcdth step of the level xkcd-5" feeds on level xkcd↑↑↑...(xkcd times)↑↑↑xkcd xkcd-5 times. "xkcdth step of the level xkcd-6" feeds on level xkcd↑↑↑...(xkcd times)↑↑↑xkcd↑↑↑...(xkcd times)↑↑↑xkcd↑↑↑...(xkcd times)↑↑...[xkcd times]...↑↑↑xkcd xkcd-6 times.

Note that I'm doing these progressions in the same way that I did the Level progressions, so this isn't also well defined, but eventually, you'll get to "the xkcdth step of the level 6", which is to be defined as the last number 9(x) as seed, that is feed on the level xkcdsomethinghuge that would take xkcd paragraphs to explain, or something 6 times. Decreasing like this until the last number 9(x) is used as seed on a higher and higher and higher and higher and higher level, 5, 4, 3, 2, and finally, once, on (higher and higher)*5 9(x)'s, until everything collapses on a number. That number is x. Use 9(x). This number will be ۹[1][1]

That's for the first part of my confusing paragraph.

"Step xkcdth-1 of the level xkcd", would then no longer start at level xkcd, it would start at the last level used to produce ۹[1][1], which I'll call level ߖ. And while Level ߖ would increase as,ߖ ߖ2, 2, ߖߖ, ߖ ^ ߖ ^ ߖ^ ( times ߖ ,( ߖ↑↑↑... Man, the ߖ character sucks when writing, I don't know where a character will end up next, but it follows the progression of above; 9(x) is no longer used, instead, at this point, 9(9(x)) is used, but it grows as below. And the times of repetitions would no longer be xkcd times, they would be 9(ߖ) times.

So, put ۹[1][1] into level ߖ, get x, call 9(9(x)), get ۹[2][1], put it into level ߖ, call 9(9(x)) get ۹[3][1]... repeat until you get to ۹[9(*)][1], this is ۶[1][1].

"Step xkcdth-1 of the level xkcd-1": Put ۶[1][1] into level 2ߖ, get x, call 9(9(9(x))), get ۶[2][1], put it into level ߖ2, repeat 9(*)-1 times call 9(9(9(x))) get ۶[3][1]. Repeat that until you get to ۶[9(**)-1][1], this is ۹[1][1].

**
ߖ2

"Step xkcdth-1 of the level xkcd-2": Put ۹[1][1] into level ߖߖ, get x, repeat 9(*)-2 times call 9(9(9(9(x)))), get ۹[2][1], put it into level ߖߖ, repeat 9(*)-2 times call 9(9(9(9(x)))) get ۹[3][1]... repeat until you get to ۹[9(**)][1], this is ۶[1][1].

**
ߖߖ

Okay, you should get the drift by now about that goes on, and this only scratches the surface.

First, I think I need to define the progression of the levels, what's wrong with this progression?

x, 2x, xx, (((x^)x^)...x times...^x)^x)^x, x↑↑↑...x times...↑↑↑x, x↑↑↑...x times...↑↑↑x↑↑↑...x times...↑↑↑x↑↑↑...x times... [x times] ...↑↑↑x

It's meant to be a progression faster than A(), or something, so every step things get a lot more enormous than last time, surely the should be a way to formalize this growth unambiguously?

Anyway, I know that even if I made a clear definition of the idea and produced a definite number, it wouldn't dare to dream catching up to the numbers posted up to page 17, I believe, but maybe something can be done about that...
Go! Go! You can do it username5243!
Cheers Marsh'n!

THANKS KARHELL!!

Deedlit
Posts: 91
Joined: Sun Mar 08, 2009 2:55 am UTC

### Re: My number is bigger!

Sorry for the late reply. The problem with finding a "formalizing the growth unambiguously" is that the sequence doesn't seem to follow any set pattern. The fourth term in the sequence seems to be an iterate of the third, but then the fifth function jumps all the way to the Ackermann function. Then the sixth function is just x [x+1] x, where [x+1] means x+1 up arrows, so it's less than an iterate of the fifth function.

Let's let f(m, n) be mth function in your sequence evaluated at n, and let f(n) = f(n, n). If I had to guess, I would say that f(5 + m, n) is less than F_{omega + m} (n) where F is the fast-growing hierarchy. So f(n) is no more than F_{omega*2} (n) in the fast-growing hierarchy.

You do a lot of evaluating at various levels, but the higher levels will completely dominate the lower levels, so we only really need to consider the highest level you evaluate. That level is about f(xkcd, xkcd), and the number that is evaluated will be less than f(f(xkcd, xkcd), f(xkcd, xkcd)) = f(f(xkcd)). You then set this equal to ۹[1][1], and perform the same procedure. The number you produce will therefore be less than f(f(f(f(xkcd)))). You call this ۶[1][1]. Each new symbol adds to f's to the upper bound, so if you have m symbols you get an upper bound of f2m(n). Note that, if f is at level omega*2, f2n(n) is at about level omega*2 + 1, just one level higher! So there is still a long way to go.

Looking at your previous post, it looks like the sequence A', B', C' function much like your sequence of symbols in your most recent post, so the mth letter will generate about f2m(n), which is at level omega*2 + 1. It's not clear from your post, but it seems like the symbols A'', B'', C'' either represent f2n + 2m(n) or they represent one level further up the hierarchy. I'll be generous and assume they are one level further up the hierarchy. I'm not sure what Z'''(Y'''(X'''...A'(xkcd 's)')... 's)... 's)... 's)... 's means, but it looks like you are diagonalizing over the 's, which would take you to omega*3, or omega*3+1 if you iterate the diagonalization. You then define a new superscript °, and n ° superscripts will get you omega*3 + n, and as we add more superscripts we will get omega*m + n. You then switch to subscripts, but it's rather confusing since you use previous symbols as subscripts, which won't work. But it seems like m kinds of subscripts will get you to omega^2 + omega*m + n.

So it looks like you are well below omega^omega, which keeps you below Actaeus's best effort, which was at level omega^omega^3.

eculc
Wet Peanut Butter
Posts: 451
Joined: Mon Jun 27, 2011 4:25 am UTC

### Re: My number is bigger!

let x = "myNumber"
limit of x as x approaches infinity.
Um, this post feels devoid of content. Good luck?
For comparison, that means that if the cabbage guy from Avatar: The Last Airbender filled up his cart with lettuce instead, it would be about a quarter of a lethal dose.

tomtom2357
Posts: 563
Joined: Tue Jul 27, 2010 8:48 am UTC

### Re: My number is bigger!

eculc wrote:let x = "myNumber"
limit of x as x approaches infinity.

Not well defined.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

WarDaft
Posts: 1583
Joined: Thu Jul 30, 2009 3:16 pm UTC

### Re: My number is bigger!

Well defined, but illegal. The limit of x as x goes to infinity is infinity. Which is banned.
All Shadow priest spells that deal Fire damage now appear green.
Big freaky cereal boxes of death.

Vytron
Posts: 429
Joined: Mon Oct 19, 2009 10:11 am UTC
Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P

### Re: My number is bigger!

Deedlit wrote:Sorry for the late reply. The problem with finding a "formalizing the growth unambiguously" is that the sequence doesn't seem to follow any set pattern.

Yes, I have to work on this, the idea is to have an exponential pattern instead of a "lineal" or "diagonal" one, so every Level+1 is much more powerful than the last one, so a jump from level 95 to level 96 would advance a hierarchy higher than what level 94 to level 95 did. If some f(xkcd, xkcd) is reached at level n, level n + 1 will contain f(f(xkcd)).

Anyway, this currently feels like a car without fuel, because without the starting pattern, the full power of the levels can't be perceived, and the way the number of recursions need to be increased isn't shown properly (which is clear from your "You then set this equal to ۹[1][1], and perform the same procedure", which I'm not because applying the same procedure on ۹[1][1] would get to ۹[xkcd][1], which is a long way from ۹[*][1], and ۶[1][1] requires ۹[9(*)][1] - every time you repeat the procedure, the times you have to repeat it increase. - Just like when we reach A', it seems to me you're jumping from A' to A'' in the way it's meant to jump from A' to B', And jumping like that from A'' wouldn't be B'', but [A'th symbol]'', a long way from [A''th symbol]''=B''. A'(xkcd 's)' means an A with xkcd number of 's, which would be the number of 's that B has, which would be the number of 's that C has... which would be the number of 's that Z has, which is A°, which I think you're greatly underestimating. Repeating this whole process gets us to [A'th symbol]°, a long way from [B'th symbol]° ... [A''th symbol]° ... [A°th symbol]°=B°, a long way from A°'...)

Anyway, I'll leave this hanging in there for now, without the starting pattern this is just fluff. That's for trying to understand what I wrote.
Go! Go! You can do it username5243!
Cheers Marsh'n!

THANKS KARHELL!!

Deedlit
Posts: 91
Joined: Sun Mar 08, 2009 2:55 am UTC

### Re: My number is bigger!

Vytron wrote:
Deedlit wrote:Sorry for the late reply. The problem with finding a "formalizing the growth unambiguously" is that the sequence doesn't seem to follow any set pattern.

Yes, I have to work on this, the idea is to have an exponential pattern instead of a "lineal" or "diagonal" one, so every Level+1 is much more powerful than the last one, so a jump from level 95 to level 96 would advance a hierarchy higher than what level 94 to level 95 did. If some f(xkcd, xkcd) is reached at level n, level n + 1 will contain f(f(xkcd)).

Anyway, this currently feels like a car without fuel, because without the starting pattern, the full power of the levels can't be perceived, and the way the number of recursions need to be increased isn't shown properly (which is clear from your "You then set this equal to ۹[1][1], and perform the same procedure", which I'm not because applying the same procedure on ۹[1][1] would get to ۹[xkcd][1], which is a long way from ۹[*][1], and ۶[1][1] requires ۹[9(*)][1] - every time you repeat the procedure, the times you have to repeat it increase.

Yes, I know. I didn't mean "perform the same procedure with xkcd", I meant "perform the same procedure, but with xkcd replaced everywhere by ۹[1][1]." That definitely wouldn't produce ۹[xkcd][1].

- Just like when we reach A', it seems to me you're jumping from A' to A'' in the way it's meant to jump from A' to B',

I assumed that if g(n) = (nth letter)', then A'' would be something like g(g(g(...g(n)...))) with A' g's. If it's even stronger than that, you'll have to spell out what it is.

And jumping like that from A'' wouldn't be B'', but [A'th symbol]'', a long way from [A''th symbol]''=B''.
[\quote]
Wait, isn't B the second symbol? How then can [A''th symbol]''=B''? This is all very confusing.

A'(xkcd 's)' means an A with xkcd number of 's, which would be the number of 's that B has, which would be the number of 's that C has... which would be the number of 's that Z has, which is A°, which I think you're greatly underestimating.

You're doing the same thing that you're doing with the levels, which is making this really long chain, when the strongest level of the chain will dominate all the others anyway. Anyway this is just diagonalizing over the number of 's, and since each ' goes one level up the fast-growing hierarchy, ° will be one omega level up the fast-growing hierarchy.

Darvince
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### Re: My number is bigger!

g(x)=f(x)+1
mine will always be bigger.
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tomtom2357
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### Re: My number is bigger!

Illegal, using previous definitions. Also, you're not actually defining a number.
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Darvince
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### Re: My number is bigger!

ok mommy
f(x)=the next number to be posted
g(x)=f(x)+1
g(x)=largest number

iwin.
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orangedragonfire
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### Re: My number is bigger!

1

Just to thwart Darvince.

Also, why do you use f(x) to represent a number? It's customary to let f(x) be a function.

Vytron
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### Re: My number is bigger!

Deedlit wrote:Yes, I know. I didn't mean "perform the same procedure with xkcd", I meant "perform the same procedure, but with xkcd replaced everywhere by ۹[1][1]."

But I'm not just replacing every xkcd for ۹[1][1], I'm actually replacing xkcd for ߖ, which is... huh, let ❶=the xkcdth step of the progression of that I have yet to define, then ߖ= ❶(((...(xkcdxkcd)((xkcd2)((2xkcd((xkcd))...))), with each parenthesis being the times you have to repeat the whole thing to remove a layer (yes, I still have to define the progression...)

I'm not just getting a new symbol and feeding it again to the loop, I'm increasing the loop length needed to get to the next symbol, and I'm increasing the times the loop is increased to get to the next symbol, and I'm increasing the times I increase the times, and I'm increasing the times I increase the times I increase the times (keep talking like this for ߖ times). And each increment is following the undefined progression.

So ۹[xkcd][1] is the upperbound of the highest number you could get if you repeated the whole process but used ۹[1][1] itself in the place it mattered the most, and if you fed ۹[xkcd][1] in the whole process (which is now much larger), you'd get ۹[2xkcd][1], because the distance from ۹[95][1] to ۹[96][1] requires a progression of repetitions much higher than what jumping from ۹[94][1] to ۹[95][1] required. So just getting a new symbol and feeding it into the whole process advances one step in the progression. Repeating the whole process (not the other whole process, but the process of repeating the whole process) ۹[1][1] times gets you to ۹[xkcd][1], the times you require to repeat the whole process to repeat the whole process to repeat (repeat as necessary) increase up exponentially, so * is much harder to reach than this, and 9(*) is 9(10x10100) orders of magnitude higher than the difference from 1000 and a 90 Googols. You need to reach ۹[9(*)][1] for ۶[1][1], and the distance to ۹[1][1] is much larger, because if you repeated the whole process (the whole process up until now) with ۶[1][1] in there, it wouldn't even reach ۶[2][1], because ۶[2][1] requires repeating the whole process (the whole process up until now) 9(*) times, while the number grows in a much faster fashion. Reading again what I wrote, it seems this is very unclear, but at each symbol the times "repeating the whole process" need to be repeated increase faster and faster.

I think the power you're assuming A' has actually should have been surpassed at about ۹[n][n][n]...xkcd [n]s...[n][n][n], so now I think once I really come back to this, your underestimation of my number would be much greater, what you think is A'' would turn out to be about ۶[n][n][n]...xkcd [n]s...[n][n][n]. The "...keep talking like this for xkcd more paragraphs..." in my original post isn't making my number justice...

I really sound like an old man babbling, without the base progression, all my attempts to explain how the number grows seem futile.
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Deedlit
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### Re: My number is bigger!

Vytron wrote:
Deedlit wrote:Yes, I know. I didn't mean "perform the same procedure with xkcd", I meant "perform the same procedure, but with xkcd replaced everywhere by ۹[1][1]."

But I'm not just replacing every xkcd for ۹[1][1], I'm actually replacing xkcd for ߖ, which is... huh, let ❶=the xkcdth step of the progression of that I have yet to define, then ߖ= ❶(((...(xkcdxkcd)((xkcd2)((2xkcd((xkcd))...))), with each parenthesis being the times you have to repeat the whole thing to remove a layer (yes, I still have to define the progression...)

You said that ߖ is the last level in the definition of ۹[1][1], so that ۹[1][1] is greater than ߖ. Your expression ❶(((...(xkcdxkcd)((xkcd2)((2xkcd((xkcd))...))) is the same process you described before, except that you go up to ❶, which is less than ۹[1][1]. So if ۹[1][1] is f(f(xkcd)), then the next number will be less than f(f(f(f(xkcd)))).

I'm not just getting a new symbol and feeding it again to the loop, I'm increasing the loop length needed to get to the next symbol, and I'm increasing the times the loop is increased to get to the next symbol, and I'm increasing the times I increase the times, and I'm increasing the times I increase the times I increase the times (keep talking like this for ߖ times). And each increment is following the undefined progression.

Yes, I know. The same is true of the fast-growing hierarchy. By using f(n) to stand for the number you get when you apply your level-growing procedure n times, it sounds like I am diminishing your notation; really it is just simplifying your notation to make it more understandable, but the effect is to make the procedure sound less grandiose. f(f(n)) is, as you say, more than just feeding it again into the loop, itg increases the loop length and all that. And f(f(f(n))) increases the loop length even more, and so on. All these functions will be dominated by fn(n) = f(f(f(...f(n)...))) with n n's, which we can call g(n). This represents one level up the fast-growing hierarchy. And then you can go to g(g(...g(n)...)), which is barely describable in terms of the original f(n). I think you would be surprised just how big gn(n) would be in terms of your original level-growing procedure. And this is just two levels up the fast-growing hierarchy.

So ۹[xkcd][1] is the upperbound of the highest number you could get if you repeated the whole process but used ۹[1][1] itself in the place it mattered the most,

۹[xkcd][1] is the number you get if you repeat the process xkcd times, but clearly if you repeat the process more you will get a higher number, right?

and if you fed ۹[xkcd][1] in the whole process (which is now much larger), you'd get ۹[2xkcd][1], because the distance from ۹[95][1] to ۹[96][1] requires a progression of repetitions much higher than what jumping from ۹[94][1] to ۹[95][1] required. So just getting a new symbol and feeding it into the whole process advances one step in the progression. Repeating the whole process (not the other whole process, but the process of repeating the whole process) ۹[1][1] times gets you to ۹[xkcd][1], the times you require to repeat the whole process to repeat the whole process to repeat (repeat as necessary) increase up exponentially, so * is much harder to reach than this, and 9(*) is 9(10x10100) orders of magnitude higher than the difference from 1000 and a 90 Googols. You need to reach ۹[9(*)][1] for ۶[1][1], and the distance to ۹[1][1] is much larger, because if you repeated the whole process (the whole process up until now) with ۶[1][1] in there, it wouldn't even reach ۶[2][1], because ۶[2][1] requires repeating the whole process (the whole process up until now) 9(*) times, while the number grows in a much faster fashion. Reading again what I wrote, it seems this is very unclear, but at each symbol the times "repeating the whole process" need to be repeated increase faster and faster.

You definitely need to be more exact. Give precise definitions of your numbers, and could you please use regular symbols instead tiny little symbols and changing colors? It is very hard to distinguish your symbols.

I think the power you're assuming A' has actually should have been surpassed at about ۹[n][n][n]...xkcd [n]s...[n][n][n], so now I think once I really come back to this, your underestimation of my number would be much greater, what you think is A'' would turn out to be about ۶[n][n][n]...xkcd [n]s...[n][n][n]. The "...keep talking like this for xkcd more paragraphs..." in my original post isn't making my number justice...

I haven't seen you talk about "۹[n][n][n]...xkcd [n]s...[n][n][n]" before, everything was with two brackets.

Anyway, do you understand the fast-growing hierarchy well enough to determine that my comparisons are wrong?

I really sound like an old man babbling, without the base progression, all my attempts to explain how the number grows seem futile.

You can simply start with a base set of functions fm(n) for the nth number of level m. Really, you're not going to get all that far with a base progression without using massive recursion, so you might as well just pick a reasonable function and go with it. It's all about the amount of recursion you're able to extend the notation with. That's why the fast-growing hierarchy starts with just f0(x) = x+1; it's good enough to get where we're going.

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### Re: My number is bigger!

Deedlit wrote:You said that ߖ is the last level in the definition of ۹[1][1], so that ۹[1][1] is greater than ߖ. Your expression ❶(((...(xkcdxkcd)((xkcd2)((2xkcd((xkcd))...))) is the same process you described before, except that you go up to ❶, which is less than ۹[1][1]. So if ۹[1][1] is f(f(xkcd)), then the next number will be less than f(f(f(f(xkcd)))).

Ahh, I see now. ۹[1][1] is so big that ۹[1][1]th level of ߖ is completely dominated by ۹[1][1]th level of ۹[1][1], and no matter how many layers of repetitions I add, ۹[۹[1][1]][1] will be much bigger than ۶[1][1], and ۹[۹[۹[1][1]][1]][1] will be MUCH bigger than ۹[1][1], so even if you were generous and made them equal, ۹[1][2] would just be ۹[۹[۹...xkcd times...۹[۹[۹[1][1]]]...]]][1] (not even ۹[۹[۹...۹[1][1] times...۹[۹[۹[1][1]]]...]]][1]). Told like this I agree this sounds a lot less grandiose.

You definitely need to be more exact. Give precise definitions of your numbers, and could you please use regular symbols instead tiny little symbols and changing colors? It is very hard to distinguish your symbols.

Sorry about that, the original idea was to make the symbols resemble a 9, and it seems I succeeded but I don't remember what was the point in doing that...

I haven't seen you talk about "۹[n][n][n]...xkcd [n]s...[n][n][n]" before, everything was with two brackets.

Oh, the same way ۹ eventually became ۹[n], and ۹[n] eventually became ۹[n][n], it would eventually get a new bracket and become ۹[n][n][n]. The idea here is that it's increasingly hard to move to ۹[n+1][n][n], increasingly increasingly harder to get to ۹[n][n+1][n], increasingly increasingly increasingly harder to get to ۹[n][n][n+1], and increasingly increasingly increasingly increasingly harder to get to ۹[n][n][n][n] (actually, the 4 "increasinglies" would be for ۹[n][n][n+2], but you get the idea).

This is what I meant with "this only scratches the surface", as the way this thing grows should surpass the fast growing hierarchy at some point. Say, ۹ with k [n]s matches the fast growing hierarchy's power, so that f(f(n)) matches with ۹ with k [n+1]s, g(n) would match with ۶ with k [n]s, gx(n) would match with ۹ with k [x]s, and this is a long way to go to ۹ with k+1 [n]s.

Since you are making a jump from ۹[1][1] to A' in your analysis, I have a feeling that your analysis of A' holds for ۹[1][1][1], so in the above k=3.

Anyway, do you understand the fast-growing hierarchy well enough to determine that my comparisons are wrong?

You tell me, if I rewrote your analysis to be like this, would it make sense?:

Spoiler:
"Looking at your previous post, it looks like the sequence ۹[1][1][1], ۹[2][1][1], ۹[3][1][1] function much like your sequence of symbols in your most recent post, so the highest number of the first bracket will generate about f2m(n), which is at level omega*2 + 1. It's not clear from your post, but it seems like the symbols ۶[1][1][1], ۹[1][1][1], ۹4th symbol[1][1][1] either represent f2n + 2m(n) or they represent one level further up the hierarchy. I'll be generous and assume they are one level further up the hierarchy. I'm not sure what ۹[1][2][1] means, but it looks like you are diagonalizing, which would take you to omega*3 (as the number of the first bracket increases), or omega*3+1 (as you keep changing symbols with a 2 in the second bracket) if you iterate the diagonalization. You then increase the second bracket to 3, and ۹[n][3][1] will get you omega*3 + n, and as we change the symbols we will get omega*m + n. You then reach ۹[1][4][1], but it's rather confusing. But it seems like ۹[m][4][1] will get you to omega^2 + omega*m + n."

Really, you're not going to get all that far with a base progression without using massive recursion

I think the way the brackets are defined have already built-in massive recursion (as, I haven't even increased the value of the third bracket), unless I just fail at understanding the fast-growing hierarchy. What's funny, if this is really the case then I wonder if I need to define the base progression and the way the levels work, that would give precise numbers but If I can match number of brackets to omega how small or big the base seeds are doesn't seem to matter because I seem to reach about the same size anyway.

Now, I'll be ditching the ۹/۶/۹ symbols (and specially the backwards 7 symbol I can't even get to work in the middle of the sentence), I promise.
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WarDaft
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### Re: My number is bigger!

Any recursion you insert by hand, compared to ordinal collapse, is rather like putting in more +1s against the very recursion you are trying to define there. It really is. In sort, counting to [imath]ω^{ω^{ω}}[/imath] is quite simply far easier than counting to [imath]ω^{ω^{ω^{ω}}}[/imath], by more than you realize. Here's an illustration of why:
Spoiler:
In order to count to [imath]ω^{ω^{ω^{ω}}}[/imath], you must first count to:
[imath]ω^{ω^{ω}}[/imath]
[imath]ω^{ω^{ω}}+ω^{ω^{ω}} = ω^{ω^{ω}}2[/imath]
[imath]ω^{ω^{ω}}3[/imath]
...
[imath]ω^{ω^{ω}}ω = ω^{ω^{ω}+1}[/imath]
[imath]ω^{ω^{ω}+1} + ω^{ω^{ω}}[/imath]
[imath]ω^{ω^{ω}+1} + ω^{ω^{ω}}2[/imath]
...
[imath]ω^{ω^{ω}+1} + ω^{ω^{ω}}ω = ω^{ω^{ω}+1} + ω^{ω^{ω}+1} + ω^{ω^{ω}+1}2[/imath]
...
[imath]ω^{ω^{ω}+1}3[/imath]
[imath]ω^{ω^{ω}+1}4[/imath]
...
[imath]ω^{ω^{ω}+1}ω = ω^{ω^{ω}+2}[/imath]
And you start getting numbers like:
[imath]ω^{ω^{ω}+7}13 + ω^{ω^{ω}+5}22 + ω^{ω^{ω}+2}3 + ω^{ω^{ω}}[/imath]
[imath]ω^{ω^{ω}+ω}4 + ω^{ω^{ω}}3[/imath]
...
For any given ordinal a < ωω we have to count up to [imath]ω^{ω^{ω}+a}[/imath] ω times to add one multiple of [imath]ω^{ω^{ω}+a+1}[/imath]. We can have numbers like:
[imath]ω^{ω^{ω}+ω^{5}3+3}4 + ω^{ω^{ω}+ω^{5}3+2}7 +ω^{ω^{ω}+ω^{5}3+2}3 + .. + ω^{ω^{ω}+ω^{5}2+7}4 + ω^{ω^{ω}+ω^{5}2+3}4 + ... + ω^{ω^{ω}+ω^{5}2+8}119 + ... ω^{ω^{ω}+ω^{4}97+8}13 + ... ... + ω^{ω^{ω}+ω^{3}53234342362345+88}7 + ... ... ... + ω^{ω^{ω}}[/imath]
... which should hopefully show up on one line, there's only one number there. The limit of what we're doing so far is not [imath]ω^{ω^{ω^{ω}}}[/imath] however... it's [imath]ω^{ω^{ω}2}[/imath].
We then have to count to [imath]ω^{ω^{ω}3}[/imath], which is even harder, then [imath]ω^{ω^{ω}4}[/imath], harder still. Then...
[imath]ω^{ω^{ω}ω} = ω^{ω^{ω + 1}}[/imath]
[imath]ω^{ω^{ω+1}ω} = ω^{ω^{ω + 2}}[/imath]
[imath]ω^{ω^{ω2}}[/imath]
[imath]ω^{ω^{ω7}}[/imath]
[imath]ω^{ω^{ωω}} = ω^{ω^{ω^{2}}}[/imath]
Aha, you're thinking, we're getting there! No, we're not. Remember what climbing up that last exponent did? This one does the same to everything I've typed above. We get numbers where a SINGLE TERM can look like this:
[imath]ω^{ω^{ω^{37}5 + ω^{36}18 + ... + ω5 + 18}3834 + ω^{ω^{37}5 + ω^{36}18 + ... + ω5 + 17}9923498 + ω^{ω^{37}5 + ω^{36}18 + ... + ω5 + 16}23412 + ... + ω^{ω^{37}5 + ω^{36}18 + ... + ω4 + 99876}123 + ...}[/imath] Those huge numbers that the exponents plagued us with counting up to [imath]ω^{ω^{ω^{2}}}[/imath] are now things we have to deal with in the exponents.

And this is just counting up to [imath]ω^{ω^{ω^{ω}}}[/imath]. Every kind of number above is now going to become an entire category of hurdles when trying to count to [imath]ω^{ω^{ω^{ω^{ω}}}}[/imath]. And remember, everwhere you see a '5' or a '17', they're actually going to be well on their way to the massive numbers you're going to get at the end. Then we have the sequence ω,[imath]ω^{ω}[/imath], [imath]ω^{ω^{ω}}[/imath], ..., the limit of which is ε0. Climbing the epsilon numbers makes climbing exponent layers look like a cake walk. You can have epsilon numbers to the power of any ordinal, including other epsilon numbers. The limit of [imath]ε_{0}[/imath], [imath]ε_{0}^{ε_{0}}[/imath], [imath]ε_{0}^{ε_{0}^{ε_{0}}}[/imath] ... (amongst a few other effectively equivalent sequences) is [imath]ε_{1}[/imath], do the same thing with [imath]ε_{1}[/imath] to get [imath]ε_{2}[/imath], where the 0, 1, 2, etc can extend to any ordinal number, including epsilons. And these are *small* ordinals. Very small. And every time we started counting up and then swapped in an omega, whatever strategy you were stacking to increase that number just got incorporated into the structure of the ordinal.

But to really grasp how big they are, work through it slowly, trying to figure out where your notation fits in.
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Vytron
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### Re: My number is bigger!

WarDaft wrote:The limit of [imath]ε_{0}[/imath], [imath]ε_{0}^{ε_{0}}[/imath], [imath]ε_{0}^{ε_{0}^{ε_{0}}}[/imath] ... (amongst a few other effectively equivalent sequences) is [imath]ε_{1}[/imath], do the same thing with [imath]ε_{1}[/imath] to get [imath]ε_{2}[/imath], where the 0, 1, 2, etc can extend to any ordinal number, including epsilons. And these are *small* ordinals. Very small. And every time we started counting up and then swapped in an omega, whatever strategy you were stacking to increase that number just got incorporated into the structure of the ordinal.

I see. However, I think my brackets already are having that power? The change of symbol before the brackets needs to get much and much higher for a bracket to change number, and a bracket needs to reach higher and higher numbers to change the symbol, and each change in the bracket and symbol to require more and more recursions, which makes it a lot more recursive. So the power of epsilon should be reached at some k brackets (If ε_{0} is reached at the bracket that has the power of the limit of ε_{0}, ε_{0}^{ε_{0}}, ε_{0}^{ε_{0}^{ε_{0}}}, the next symbol would be the limit of ε_{x}, and an extra bracket would be the limit of ε_{ε_{ε_{ε_{...}}}}), and I don't see why it can't be reached at 5 brackets.

So, apparently, the original base progression doesn't need to be defined, because Deedlit was able to analyze my numbers without it.

Instead of the hard to read progression ۹/۶/۹, I'll just use blue numbers. So that ۹=1, ۶=2, ۹=3.

1 is the base seed, which is just a 9 followed by xkcd 9s, which is feed into some kind of hydra function that feeds on 9s on every level, and returns 9(x) (which turns it into a number that only have 9s and has x number of them) which is feed into the next level. Every level adds a new hierarchy of recursion, and iterative recursion, but the next level adds twice that, and the next adds the number of recursions exponentiated, with each new level increasing the level of recursions following the undefined progression.

Eventually, you'll reach level 1, and the limit of that number will be a, which is the new seed. The process of above is repeated, but instead of adding one hierarchy, and one iterated recursion, at every step, you add 1 hierarchy, and 1 iterated recursion, in the next level, 2*1 for both, in the next, 12, in the next 11, and so on.

At level a, the limit of that number will be b, and you repeat the process but adding a hierarchies of iterated recursions, and increasing them as above. You get c, d, e... There will be a 1th symbol, let's call it ζ. You keep repeating the recursions until you reach the ζth symbol. That's 2.

2 increases of the same way of above, but if you used it as seed and repeated the whole process changing the 1s for 2s, you would only reach the next level. So now the distance from 1 to 2 is the distance from 2 to level 1 2, and the distance to the next level grows at double the magnitude as above. To reach 2a, you need twice the distance of levels as before, to get 2b. You get 2c, 2d, 2e... There will be a 2th symbol, which we'll call 2ζ, you keep repeating the recursions until you get to the 2ζth symbol, which we can call 2ζ1, and you keep repeating the recursions until you reach the 2ζ1th symbol, which will be 2ζ2. You keep going like this until you reach 2ζ2ζ. That will be 3.

If there was a function to get from 1 to 3, and it was used on 3, you'd get a number that would be half the number you would get if you put 3 at the first level. And the distance from 3 at first level to 3 at second level would be the magnitude squared of above. The number progressions go as before, but the symbol needed to be reached for 4 would be 3ζ3ζ...3ζ3ζ3ζ3ζ times...3ζ3ζ.

The distances between the blue numbers should keep growing like this.

The highest level evaluated would be 1[1], which, if I'm understanding Deedlit's analysis right, should be equal to the power of ω. Then, to reach 1[2], we need to use it as seed and repeat the whole process increasingly more times, so that when 1[1[ζ]] is reached, that's equal to 2[1], which would be about 2ω.

This keeps being repeated, with the distances of the number in the brackets increasing all the times, and the number required in the first bracket to be much higher for a symbol change, until some ζ[1] is reached, which will be 1[1][1], or about ω2.

Does my progression make sense up to now? I know it looks very slow, but I want to know if at least it's understandable to this point. At least, instead of three brackets to get to 2ω+1, I'm using 2 to get to ω2.
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Darvince
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### Re: My number is bigger!

Wait, so if I've got this right, then at 2*omega or whatever you're already at like 10^100 digits?
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Vytron
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### Re: My number is bigger!

No, I have surpassed that with 1, considering it's powered by xkcd, which is A(g64, g64), 9(10^100) (of my 9(x) function) is much lower than g64, as is 9(10^100^100...64 towers...^100), anyway.

With 1 I just wanted to start with a huge seed, because the seed becomes the number of hierarchic recursions at the second symbol, but 1 was defeated by Mouffles at the second page (heh, he beat 1[1][1]as well).

A number with 10^100 digits would have been beaten at the first page.
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WarDaft
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### Re: My number is bigger!

I see. However, I think my brackets already are having that power? The change of symbol before the brackets needs to get much and much higher for a bracket to change number, and a bracket needs to reach higher and higher numbers to change the symbol, and each change in the bracket and symbol to require more and more recursions, which makes it a lot more recursive. So the power of epsilon should be reached at some k brackets (If ε_{0} is reached at the bracket that has the power of the limit of ε_{0}, ε_{0}^{ε_{0}}, ε_{0}^{ε_{0}^{ε_{0}}}, the next symbol would be the limit of ε_{x}, and an extra bracket would be the limit of ε_{ε_{ε_{ε_{...}}}}), and I don't see why it can't be reached at 5 brackets.

That's not quite how it works. If you have some pattern which grows as the limit of ε_{0}, ε_{1}, ε_{2}, then the next layer of recursion you add onto that is ε_{ω}. The whole thing I described about counting from ω^^3 to ω^^4 (to shortcut notation because that's a lot of brackets otherwise) is now how many 'layers' per sey, of the recursion stacking that got you to ε_{ω} that you have to do to get to ε_{ω^^4}. And even if you reach the point where you're equivalent to ε_{ε_{ε_{ε_{...}}}}), ... it's so much easier to define bigger ordinals than to specifically describe more effective recursion. Though I'm probably going to side with Deedlit on your notation being somewhere in the low ω^^2 ish region. If you're repeating your descriptions, you're probably not actually defining numbers that particularly huge.

Wait, so if I've got this right, then at 2*omega or whatever you're already at like 10^100 digits?
In the fast growing hierarchy, you surpass that size number at somewhere around F4(3).
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### Re: My number is bigger!

WarDaft wrote:Though I'm probably going to side with Deedlit on your notation being somewhere in the low ω^^2 ish region. If you're repeating your descriptions, you're probably not actually defining numbers that particularly huge.

That's because I have used a few number of brackets, but I claim I should be able to reach the power and surpass the power of any number in the thread (at least up to page 17) just by adding more and more of them, so that when A' is actually reached, it'll be particularly huge.

Stop me at the point it no longer makes sense (ζ stands for the limit of the function, which gets bigger every time):

1[1][1] ≈ {ω}2
1[2][1] ≈ {2ω}2
1[ζ][1] ≈ 2[1][1] ≈ {ω}ω
2[2][1] ≈ {{ω}ω}ω
2[3][1] ≈ {{{ω}ω}ω}ω
2[ζ][1] ≈ 3[1][1] ≈ {ω}↑↑ω
3[2][1] ≈ {ω}↑↑↑ω
3[3][1] ≈ {ω}↑↑↑↑ω
3[ζ][1] ≈ 4[1][1] ≈ {ω}↑↑↑...ω ↑s...↑↑↑ω
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### Re: My number is bigger!

okay then is 2*omega A(g64, g64)^A(g64, g64) digits?
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WarDaft
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### Re: My number is bigger!

You haven't described anything that would make just increasing those numbers give you such fundamentally more powerful recursion. You're also still not getting very far. There's no evidence that the process you're describing would continue to grow at the rate I'm about to credit it with, but even if it did, the limit would be ζ0. Which is still a small ordinal, compared to most in the thread.

Ordinals really are extraordinarily recursive in nature. So much so that you don't actually need to use recursion your actual function in the long run. You can make your function f0(x) = x+1, fa+1(x) = fa(x) + 1, fa(x) = fa[x](x). Note that x never increases here. But yet, there are ordinals sufficiently large that this catches up to the same growth rate as the fast growing hierarchy, which has definitons f0(x) = x+1, fa+1(x) = fax(x), fa(x) = fa[x](x). If these don't sound very different, tell me what the rough scale of [imath]f_{ε_{0}+ω}(4)[/imath] is in each. Use 0,1,ω,ωω... as your fundamental sequence for ε0, and 0,1,2,3... for ω.

Darvince wrote:okay then is 2*omega A(g64, g64)^A(g64, g64) digits?

That's around Fω+1(65)
Technically, you should write ω*2 rather than 2*ω (which is actually just ω), but that's not terribly important for this, we know what you mean.
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### Re: My number is bigger!

WarDaft wrote:You haven't described anything that would make just increasing those numbers give you such fundamentally more powerful recursion.

It's because it would take several pages to write, and it's meant to be deduced from the blue numbers progression.

There's an original base progression similar to the one depicted in this post, to which each biggest number is feed again increasingly more times (with the number of times increasing as the number of the first bracket increases), increasingly more times (as the blue number increases)... ... ... (increasingly more times which themselves increase as these x[x] does) (...) (...) (...) [which themselves increase the biggest number times...]

This produces the new level of recursions a, b, c, d, e...ζ...ζth symbol, which increase the blue numbers by one, but, this time the original base progression gains this power, so, feeding this into it all over again is level 1 of the levels, which is feed into itself much more times than before (the magnitude difference between 1 and 2, times... (times...) [times...] {times...})

This all isn't well defined, but I haven't bothered because from what I understood from Deedlit, doing all these recursions will not get me past ω. So the limit of doing this, which would be ζ or 1[1], would be ω.

Now, 1[1] is being used itself to increase the level of recursions, so the power of ω is gained in the original base progression. 1[1]a, 1[1]b, 1[1]c, 1[1]d, 1[1]e... would contain ω + n, so that 1[1]ζ = ω + ω = 2ω = 1[2].

And now the original base progression gains the power of 2ω, so 2[1], 3[1], 4[1] contain nω, so that ζ[1] = ω * ω = ω2 = 1[1][1].

Now, the original base progression power of ω2 is gained, and the number of times the recursions are added become exponential. So, 1[1][1]a, 1[1][1]b, 1[1][1]c, will contain ω2 + n, ω2 + ω + n, ω2 + 2ω + n, ω2 + nω + n, so that 1[1][1]ζ = ω2 + ω2 = 2ω2 = 1[2][1].

However, now the distance from 1[1][1] to 1[2][1] will be the first level of the original base progression, so 1[3][1], 1[4][1], 1[5][1] will contain ωn, so that 1[ζ][1] = ωω = 2[1][1].

Now the original base progression gains the power of ωω, so that the number of times 2[1][1] needs to be feed into itself to reach 2[1][1]a needs to be high enough that we count to 2ωω, 2[1][1]b to ωω+1, 2[1][1]c to ω, 2[1][1]d to ωω^2, so that 2[1][1]ζ = {{ω}ω}ω = 2[2][1]

Now the original base progression gains the power of {{ω}ω}ω...

WarDaft wrote:You're also still not getting very far. There's no evidence that the process you're describing would continue to grow at the rate I'm about to credit it with, but even if it did, the limit would be ζ0. Which is still a small ordinal, compared to most in the thread.

But you're talking about the limit of ζ[1][1], right? Also, I think that'd be the limit of ω[1][1], which, if we used blue {}'s to describe the xth blue symbol of the value, would just be {1[1][1]}[1][1]. ζ[1][1] would actually be ζ1 = 1[1][2].

Otherwise, it doesn't make sense to talk about ζ, because the limit of the function grows indefinitely, so the ζ in 1[ζ] is much more smaller than the ζ in 2[ζ].

WarDaft wrote: Use 0,1,ω,ωω... as your fundamental sequence for ε0, and 0,1,2,3... for ω.

If it isn't clear enough yet, the main idea with my number is that whatever growth x+1[x][x][x] shows, will be gained by y[x+1][x][x], and when you reach the limit of the blue digit, you get x[x][x+1][x], which becomes the new growth of x[x+1][y][x]. It may be confusing because the number of the first bracket is increased before the blue number can increase, but I'd rather have it like this because the next level of hierarchies can clealy be seen in the next blue digit.

If you meant to turn my original base progression into that fundamental sequence, I think it'd get so big that I wouldn't be able to compare it, because, I don't even know what's the xkcdth level of that sequence.

I also don't know if using 0,1,ω,ωω instead of 0,1,ω,2ω,ω2ω would help the internal sequences, as that would just require to use less symbols, but I'm not spelling out all of them anyway.
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### Re: My number is bigger!

The sequence 0, 1, ω, ωω, is the iteration of the function f(x) = ωx starting at 0. ω0 = 1, ω1 = ω, etc. You can use any strictly increasing countable sequence of ordinals less than ε0 with order type ω, so long as for every ordinal less than ε0 is less than some ordinal in your sequence, but these can be arbitrarily difficult to compute, and without using a simple relation, you can make fε_0(3) arbitrarily large, or even arbitrarily small.

Also, every time you take the xkcdth level of something, you're making your function constant. Which is bad.

And no, ζ0 is the already named limit of the sequence 0, ε0, εε_(0), εε_(ε_(0)), εε_(ε_(ε_(0)))... each of which is far more recursive than the last. These are not trivial jumps, in the slightest.

The levels you describe in your previous post are minor at best up to level 5, at which point you obtain growth rate just below fω+1(n). At which point the xkcd number becomes irrelevant. Every single time thereafter that you use it, you're shooting yourself in the foot. Furthermore, you've defined no clear progression past level five. You've gone from repeated exponentiation at level four to repeated up-arrows at five to... something else at level six. There's no clear growth rate to extrapolate to higher levels. The recursive rate you define in your paragraphs is odd, but also not that much either. Suppose item x of level x in your first heirarchy is <= fa(x) for some ordinal a (which will exist). Then if we let a_1 = the xcd'th item of the xkcd'th level, a_2 = the a_1th item of the a_1th level, etc, then a_n = fa+1(n). Note that this is more powerful than what you did in your first paragraph, because we started at the xkcd'th level and iterated up by the number we most recently reached n times, rather than down iterated down a constant number of times. Most of the rest of what you said in that post is overshadowed by this one thing.
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### Re: My number is bigger!

i like how this thread has turned into recursing A(g64, g64)^A(g64, g64) enough times that the number seems to approach infinity
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### Re: My number is bigger!

No it doesn't. Compared to infinity, every finite number, even the largest in this thread, is effectively indistinguishable from zero.
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### Re: My number is bigger!

I dunno about zero. That might be going too far. How about seven?
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WarDaft
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### Re: My number is bigger!

Well, if we had infinitesimals, that would do, but I don't think we've bothered introducing them to this thread yet... kind-a haven't seemed useful.
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### Re: My number is bigger!

WarDaft wrote:Also, every time you take the xkcdth level of something, you're making your function constant. Which is bad.

Oh, I stopped doing that long ago. Originally, I was using 3 different counters, that increased in three different ways, each every time more powerful than the last. However, Deedlit made me realize that one of the counters was completely dominating the other ones, so that x[ [x][1] ][1] was already dominating any number of recursions that I was having on the second bracket.

From that point out I stopped stopping at the xkcdth level and going up to the highest number discovered level and going to higher and higher levels than it (say, if one goes up to level 1[1], the next will go up to level 2[1]*2, the next to level 3[1]2, the next to level 4[1]4[1]... so that when 1[1][1] is reached, it would already be going to level 1[1][1]↑1[1][1]), though, in my recent re-tellings of the story, I haven't bothered to explain at what level is stopped, because the times the number is feed into itself, and the number of recursions and recursive hierarchies that are added into it are completely dominating any reached level. Multiply 2*2 enough times and it will reach ω at some point, and the number of multiplications will completely dominate if you started with that or 2*2.1.

WarDaft wrote:And no, ζ0 is the already named limit of the sequence 0, ε0, εε_(0), εε_(ε_(0)), εε_(ε_(ε_(0)))... each of which is far more recursive than the last. These are not trivial jumps, in the slightest.

Ah, my bad, I was using the ζ symbol for something else... That's why originally I was using ۹/۶, unused symbols but the problem with them was they looked too similar. Oh well, let's use ¶ as the limit of the function.

¶ grows in this manner:

We have n[n][n]. For n to change, you need to reach n[¶][n], which will be n+1[n][n]. However, by that point ¶ will be n+1[n][n], so any kind of growth you'd expect from adding to n, will now happen when adding to o[n+1][n], which now requires a much higher ¶ to advance. So, it doesn't matter that the jump from ε0 to εε_(0) is "non trivial", because if you were able to jump from ω to ε0, and from ε0 to ε1, then, at some point in the brackets, the "non" trivial jumps will be trivial.

Let's go back to:

ω[1][1] = {1[1][1]}[1][1], and say that by this point, things are so recursive that ε0 is reached.

By this point there are a lot of recursions going on. First, there's the original level progression that advances with the power of ε0 in the levels. Then, there's the ω[1][1]a, ω[1][1]b, ω[1][1]c... sequence, that is adding a new level of ε0 hierarchies and ε0 levels of levels, and ε0 {levels of levels of levels...} with each progression.

Then there's {1[1][1]}[n+1][1], which keeps doing all this until ω+1[1][1] is reached. However, by this point, what you expect to be ω+2[1][1] (if everything else was repeated), would actually be reached as soon as ω+1[2][1] is reached. So adding to ω becomes harder and harder.

Then there's also {1[n+1][1]}[1][1], following the same rules, and {n+1[1][1]}[1][1], following the same rules.

Eventually [1][1]}[1][1] is reached, or, {{1[1][1]}[1][1]}[1][1], which would be εω

Now, you may think that if everything was repeated again, {{ω[1][1]}[1][1]}[1][1] would be reached, and it would be just εω+1. However, the original progression already has the power of εω, so that εω+1 is reached at {{1[1][1]}[1][1]}[2][1]. εω2 would be reached at {{1[1][1]}[1][1]+1}[1][1]. εω3 at {{1[1][1]}[1][1]+1}[2][1], εω^2 at {{1[1][1]}[1][1]+2}[1][1], εω^3 at {{1[1][1]}[1][1]+2}[2][1], εω^ω at {{1[1][1]}[1][1]+3}[1][1], ε{{ω}^ω}^ω at {{1[1][1]}[1][1]+3}[2][1], and so on.

By this point, the level of recursions would be higher, so there's {{1[1][1]}[n+1][1]}[1][1], {{1[1][1]+1}[1][1]}[1][1], {{1[n+1][1]}[1][1]}[1][1], {{n+1[1][1]}[1][1]}[1][1], and the sequence:

{{1[1][1]}[1][1]}[1][1]
{{{1[1][1]}[1][1]}[1][1]}[1][1]
{{{{1[1][1]}[1][1]}[1][1]}[1][1]}[1][1]

Eventually you reach

{{... ω[1][1] {'s...{{1[1][1]}... ω[1][1] }[1][1]'s...}[1][1]

AKA

[1][1]

Or

1[1][2] = εε_(0)

And now, you could say, "if you repeated all that, then, at 1[1][¶] you would reach εε_(ε_(0)))", or something. However, the recursive power of εε_(0) is gained at the level progressions, so that εε_(ε_(0)) is reached at 1[2][2], εε_(ε_(ε_(0))) at 1[3][2], εε_(ε_(ε_(ε_(0)))) at 1[4][2], and so on.

WarDaft wrote:There's no clear growth rate to extrapolate to higher levels.

There's no clear growth rate at the original levels either. But, I think it doesn't matter, because when analyzing the numbers, if I managed to count to ω with the original undefined progression, then I just need to add layers of recursion using the number that counted to ω, to reach ω+1, and do that enough times to reach ω+ω, and do THAT enough times to reach ω*ω, and so on.

Suppose my progression was off the mark, and that what I call the 1[1][1], 2[1][1], 3[1][1] progression couldn't be attained, and it required some 1[1][1], 10[1][1], 100[1][1], 1000[1][1]... progression. I don't think it would matter, because the way to access to ε0 would remain the same, because these number are so far away from ω that all it matters is how fast ¶ grows.

¶ looks at how fast the number moved since the last time a change happened (like going from 1[1][1] to 1[1][2]), and then becomes so big that for accessing the next blue number, the first bracket has to reach the limit of the sequence of that change.

So 1[¶][2], or 2[1][2], would reach ζ0. 2[2][2] would access the next ordinal in the sequence ω0, ε0, ζ0... 3[1][2] would reach the limit of the ordinals, and 3[2][2] would access the limit of the limit. If you think this is impossible, I'll be happy to move these powers to 1[1][3], 1[1][4], 1[1][5] or even to 1[1][1][1], 1[1][1][1][1], 1[1][1][1][1][1], but remember that whatever power 1 with k [1]'s has, will be gained, and instead of having to add an extra bracket to reach the next limit, you do that by adding 1 to the first bracket.

So I think my mistake was linking to my original post where I was sticking to xkcd levels, because my number has gotten much more powerful since then, I think.
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### Re: My number is bigger!

Hey guys. I wanted to try creating my own function to create extremely large numbers. I know it won't beat the numbers now, but I just want to know how big you think my number compared to the others in this thread. Note that this function features the use of Tricky's Extended Ackermann Function

Define P(n):
P(n) = A(n, n, n, n, n, n.....,n ,n) with "n" occurences of n
P(0) = 1
P(1) = 2

So: P(2) = A(2, 2) and P(3) = A(3, 3, 3)
This is fairly simple and would create rather small numbers. So let's up the ammunition

Define Pm(n):
Pm(n) = A(P(n), P(n), P(n).....P(n), P(n)) with "m" occurences of P(n)
P0(n) = P(n)
P1(n) = P(n) + 1

Therefore: P2 (2) = A(P(2), P(2)) which in turn equals A(A(2, 2), A(2, 2))

Now define Pq (n):
Pq(n) = P(P(P(P(P(P(.....P(n)))))) with "q" sets of brackets
P0(n) = P(n)
P1(n) = P(n) + 1
Pmq(n) = Pm(Pm(Pm....(n))))) with "q" occurences of n

Here we can see that this grows quite large quite quickly.
As an example: P22(2) = P2(P2(2))
This then results in: P2(A(A(2, 2), A(2, 2)) = P2(A(7, 7))
Now A(7, 7) is already a large numbers, but there's still a bracket left to deal with!
The result of this would be A(P(A(7,7)), P(A(7,7)) and P(A(7,7)) is the ackermann function with A(7,7) occurences of A(7,7)!
As you can see, this is huge, with a large m, n, and q

But, I'm not done yet.

Define &(n):
&(n) = Pnn(n)

Also define &m(n):
&m(n) = &(&(&(&(&(&(&(&(...(n)))))))... with "m" sets of brackets

And define &q(n):
&q(n) = &q(P(A(n, n, n....n)) with "q" occurences of n
&mq(n) = &&m(P(A(n, n, n..., n))(P(A(n, n, n, n...n)) with "q" occurences of n

Using this last function, I can create this monstrous number which I call my own.
&&xkcd(xkcd)&xkcd(xkcd)(xkcd)
I wonder how monstrous this number is...