xkcd

WarDaft

The largest uncontested number is currently Deedlit's. There are two potentially larger numbers, but they are floating around at sizes where you can't hand-wave things that are being hand-waved, and I'll let someone with a sufficient math degree suss that out. Instead, let's add more chaos. I'm goin...

WarDaft

I try to read the contents of the pile out loud, accidentally completing the ritual.

Arrrrr matey, thar be a C'Thulu in the pile!

WarDaft

Well, in the interest of keeping things rolling...

Since {ZF}0(n) grows faster than BB_a(n) for any recursively defined, naturally the next stop on the crazy train is {ZF}[ZF|{ZF}₀(10¹⁰)](10)

WarDaft

Oh, yes, that's defnitely bigger. The thing was you used one on an older number that had a 10 in it rather than an 11, and you applied it to the final value, which wasn't enough to overcome the bump to 11.

WarDaft

Aha, the oracle machines show up. Oracle is the word your looking for, when you give a Turing Machine the ability to solve the halting problem. I'm very sorry Vytron, but that's actually smaller than {ZF+Con(ZF)} 11 (10). {ZF} 10 (10) already includes statements which describe oracle Turing machines...

WarDaft

{ZF} ω^77 (77↑ 77 77) (I personally feel that applying these simple recursions to the ZF thing is actually far more naïve than saying "Graham's plus one" - relatively speaking. I wonder if there's an objective way to compare such things on such vastly different scales) In a sense, I agree...

WarDaft

Well escalated quickly. Or, you know, not. As posted, username5243's number is undefined, because there is no largest number not definable in a finite number of characters in ZF. As far as emlightened, I don't actually know how many characters it takes to express the operation of a Turing Machine in...

WarDaft

'Ello. This version is not for the faint of heart. There are three rules. 1. Unless it is blatantly obvious, you must show that your number is finite when you post it. 2. Unless it is blatantly obvious, you must show that your number is winning, otherwise it is not. 3. You must define a specific num...

WarDaft

Okay, that all makes sense for the multi play case.

But what about the single play case? Does it make sense to play the game once, or no times at all?

WarDaft

So, you're dying, and the reaper shows up early to play a game with you. The game is, the reaper will flip a biased coin. There is a 55% chance of heads, which will double your remaining earthly time, and a 45% chance of tails, which will reduce your remaining earthly time by 70%. The reaper will pl...

WarDaft

I marvel at the density of the pennies, amazed at how a pile with an average thickness of 1.15 pennies managed to crush two adult humans. In my awestruck state, I accidentally take the ability to get the reference and put it in the pile. I take a penny. In a large penny on the penny, there is a diso...

WarDaft

Fats
Parry?

WarDaft

Suggesting any of these as the worst movie ever when there are movies in IMDB with a rating of less than 2 is... really rather quite mean.

WarDaft

Specifically, you may use the operations /,-,+,*,^,BB(), grouping via parentheses, and the value 1; where BB(n) is the maximum number of shifts of a 2-symbol n-state Turing Machine.

Here, let me get things started:

BB(1)/(1+1+1+1+1+1)

WarDaft

Where BB(n) is the maximum number of shifts of a 2-symbol n-state Turing Machine:

BB(3)^2 = 11449

WarDaft

Very well then, n|0 = n n|X = n+1|X-- (b)-- = b (X,0,0)-- where X contains no non-zero terms = <(X,>0<)> (X,0,b,Y)-- where X contains no non-zero terms = <(X,>0<,b--,Y)> (a,X,0,b,Y)-- where X contains no non-zero terms, and b does not contain square brackets = <(a--,X,>((a--,X,0,b,Y))<,b--,Y)> (a,b,...

WarDaft

Vytron wrote:People watching know that the notation can reach:

n,{0-0} = φ(1,0)
n,{0,0-0} = φ(2,0)
n,{{0}-0} = φ(ω,0)
n,{{{0}-0}-0} = φ(φ(ω,0))
n,{0-0-0} = φ(1,0,0)

The way you do that is by basically re-inventing the Veblen notation in your own words. There's not really any way around that.

WarDaft

Vytron, what, exactly is the reduction for 3{0-{0,0-0}}, step by step? Just because {0-X} is a more powerful function in general does not mean it is always larger than X,0, for the same X in both cases. If 3{0-{0,0-0}} does not become 3{0-{0-{0-{0-{0}}}}, what does it become? And I've already said t...

WarDaft

10^10^10^10

A discrete 10-hyper-hyper-hyper-cube.

WarDaft

Don't take my accepting of that as meaning it was true, merely that I accepted it without further fuss. In my own notation, (((0,1))) is larger than ((0,1),0), even though the latter is applying a more powerful function [ ( ,0) instead of (( ))] to the same base value. The difference is in what is r...

WarDaft

Deedlit wrote:I know WarDaft could really dial things up as well if he wanted to.

That is indeed what the colon is for.

WarDaft

Wow, I honestly don't know how I managed to fail at copy and pasting . n|0 = n n|X = n+1|X-- (b)-- = b (X,0,0)-- where X contains no non-zero terms = <(X,>0<)> (X,0,b,Y)-- where X contains no non-zero terms = <(X,>0<,b--,Y)> (a,X,0,b,Y)-- where X contains no non-zero terms, and b does not contain sq...

WarDaft

Hmm, that actually does look like the appropriate rule. It would seem I am missing a base case, because that was not the rule I was applying in my head. Let me think... this might not actually be a problem in the long run though it would change the nature of the example proof. (0,1) would be the fir...

WarDaft

Ah, yes, that one should read (0[b:]0,X)-- = <(>0<[b--:]0,X)>, I moved the positional term from after the square braces to the first place in them, and that line of definition got messed up in the update. I'll double check the rest. The updated rules. n|0 = n n|X = n+1|X-- (b)-- = b (X,0,0)-- where ...

WarDaft

< > is a repetition operator, so I don't have to specify with a function or keep typing "(0,0,..0) for n 0's" 5|(0,0) uses the rule (X,0,0)-- = <(X,>0<)> , X is an empty array, so <(X,> is really just <(>, so it simplifies to 5|(0,0) = 5+1|<(>0<)> = 6|(((((0)))))

WarDaft

I generally agree with you up to n,{0,0} = ω^ω, with caveats about being careful about making sure your notation terminates properly because I'm not going to prove that... But then: n,0,{0,0} = ω^ω*ω = ω^(ω+1) n,0,0,{0,0} = ω^(ω+1)*ω = ω^(ω+2) n,{0},{0,0} = ω^ω*ω^ω = ω^(ω*2) n,0,{0},{0,0} = ω^(ω*2)*...

WarDaft

Okay Vytron, here goes, I'll try not to skip anything. So, only two parts of my notation were used so far. First off, every array enclosed in parenthesis acts more or less exactly like an ordinal. Second, this will be based on the Hardy Hierarchy. Third, a[n] will refer to the n-th item from the fun...

WarDaft

We do not know that BB(10) is less than g2.

We do not have concrete values for anything larger than BB(4). BB(10) could be larger than every number in the serious bigger number thread.

There is no possible way to approximate the BB function, at all. It is literally not something you can do.

WarDaft

I'm cleaning it up, and building a longer, more detailed explanation in another browser tab. 'n' is indeed shorthand for <(>0<)>, because it's far more readable, they are functionally equivalent.

If it's causing errors, I guess I'll switch back to colons rather than semi-colons.

WarDaft

Well

Spoiler:

WarDaft

Technically, every recursive notation is really just a mapping to ordinals. That doesn't make them all easily readable, and there may be new errors introduced with a new format that we don't notice at first. I'll type up something later about how my notation works, and what is required when dealing ...

WarDaft

It started to feel like a Sisyphean task to watchdog Vytron's notation, find something wrong with it and he starts over, negating any work you've done to understand everything so far.

As for no one challenging my claims... there's not a lot I can do about that.

WarDaft

Here's an idea vytron. Why dont you fix, JUST the mistakes that people point out, instead of discarding everything you just did everytime someone quotes a portion of your text? Do you really expect me to make the effort to learn how your stuff works every other day? Because that is sort of the reas...

WarDaft

The way it can jump like that is because Ω (just Ω, no Ω^Ω or anything) is bigger than every ordinal it is used to define. ψ(a) = φ(1,a) for a up to φ(2,0) where ψ(φ(2,0)) = φ(1,φ(2,0)) = φ(2,0), but it stops here because the ψ function can't make zetas on it's own, so it can't 'use' ψ(2,0) when it'...

WarDaft

Maybe I'm missing something, but I don't see any rules that would make two adjacent {..}+ terms multiply their ordinal value together. That's just a weird operation to have. Considering that you yourself are getting two different results... I call shenanigans . Also, ψ(Ω) = φ(2,0), if you're meaning...

WarDaft

Hmm, I think I also just noticed that I got part of my notation switched around from what I originally intended to go with. I'll fix it when it becomes relevant.

WarDaft

No, there aren't any gaps in the countable ordinals constructed (unless you choose omega to be countable, but that's just nitpicking) P(W^W+W) is indeed phi_2(G_0+1) P(W^W+W*2) is phi_2(G_0+2) P(W^W+W*a) is phi_2(G_0+a) The fixed points of phi_2(G_0+a) is phi_3(G_0+1) P(W^W+W^2) is phi_3(G_0+1) and ...

WarDaft

Okay, time to drag things into the 21st century! < > are the repeater operator, such that <X> means the string X repeated n times ( ) denote an array, which is to be evaluated lazily, rather than immediately. [;] is an advanced delimiter. : and ; will not work the same, so I'm not re-using :. Capita...

WarDaft

That's one of the reasons I didn't nitpick the notation too much. Using the hyphen notation correctly, it would be the only thing that mattered and the end target would still be reached. I suppose I should have checked that the hyphen notation was working optimally.

WarDaft

I'm going to take this time to introduce another notation I've been pondering for ages but never really explored. This one doesn't even use numbers, just parenthesis. (a,X,[Y],Z)-- = <(a--,X,(a--,X,[Y],Z)> (a,X)-- = <(a--,X)> [ ]-- = () [a,X]-- = <[a--,X]> ()-- = null, or nothing. I have called it t...

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