TREE(3) is indistinguishable from zero compared to Loader's number[1], which basically takes every bit pattern up to some n and expresses this as a program in the Calculus of Constructions. This system is a bit weaker than being Turing complete, but the programs do always terminate (this makes the number computable compared with, say, the Busy Beaver number which does a similar thing with Turing complete programs).
It also has the geek cred of being represented by an obfuscated C program (the unobfuscated verson is also available[2]).
What you've made here is what's known in googology circles as a 'salad number'. It doesn't make a conceptual leap above what the TREE function can do anyway and is kind of similar to saying TREE(3) + 1.
SCG(13) is much bigger than your K(3) and Loader's number is much bigger than SCG(13). There are (computable) numbers bigger than Loader's number which are mentioned in the googology link in my previous post (Greedy clicque sequences and finite promise games).
> But it's the m(x) that's actually large in that example.
Sure, it's large (mindblowingly so), although I'm not quite clear exactly how you define m(x).
It's hard to get a grasp on just how much vastly larger each of these 'rockstar' numbers is to the previous one when you ascend this conceptual hierachy. (To a certain extent, that's the point of the article.)
You could make a K(K(K(...(n)...) function where you painted a K on each particle in the visible universe (about 10^80) and it still wouldn't come close to Loader's number. The point is that you can't simply recurse TREE and hope to get a number that's conceptually any bigger than TREE itself.
Yea, at some point you really can't tell if function 1 grows faster than function 2. Anyway to pick a large m(x).
m(0) = 1; m(x) is the recursion of all finite functions defined using k(m(x-1)) symbols that produce larger numbers than their input defined at stage m(x-1).
So a 1 step function F1(x) = X + X and F2(x) = X ^ X,... including any function that has been included in a mathematical paper to this point that takes one ore more finite numbers as an input.
And because these are all finite functions call them recursively in whatever order produces the largest output starting with the initial input (x). Now, to size this depends on what initial notations are allowed, so m(1) is a finite number.
Next level could then include F#(x) where F1(F2(x)) etc.
PS: The above is roughly equivalent to saying infinity +1 and is not interesting IMO. I can also drop the k from the above, but again the point is to use the dumbest approach that works.
Sounds a bit like Yudkowsky's number[1] (your main step is the first one, everything else is salad), which is a computable version of Rayo's number[2])
I think the distinction is that Graham's Number is a "useful" number; i.e. it was used to help answer a problem. As you note, G + 1 is larger than G but G + 1 has (so far) no known significance.
BTW, I totally geeked out reading about Graham's Number. Thanks to the OP!
As the article states, it wasn't actually used in the proof where it's credited. Graham later cited it because it was easier to describe than his actual upper bounds.
I'm a big fan of the busy beaver family of problems and the halting problem is particularly interesting.
Naturally, the definition of the numbers and thought process used to describe them is orders of magnitude more interesting than the numbers themselves.
yeah but +1 is interesting if applied to the argument, not the function. TREE(3+1) instead of TREE(3)+1. In essence, +1 is all you have to work with if you want to stay in a sound logic, the question is just where to apply it. The logical conclusion would be op[TREE+1](3) in array notation.
Ah, I was wondering where the Busy Beaver idea came into this. The computability thing makes sense, ie. we can define a number using Busy Beavers but we can't ever really even reason about how big it is.
I don't think that's a helpful way to think about the Busy Beaver or non-computability. We know S(2,2) is exactly six, even though the function S is non-computable.
The tricky bit isn't running the ones that halt and noticing when, but reasoning about all the ones that _don't_ halt in order to convince ourselves that they really never halt. The larger and more complex the machine, the harder it may be for a human to reason about it in this way.
For example, suppose it's 1985 and you are examining a machine which seems to be checking all integers X above 204 to see if it can find a^X + b^X = c^X. Does it halt? Unfortunately the only way to answer that question is to wait until 1996 when a proof is published which says Fermat was right, there are no values of X for which this is true. So the machine never halts.
[Edited to try to get HN to format the question properly]
It also has the geek cred of being represented by an obfuscated C program (the unobfuscated verson is also available[2]).
[1] http://googology.wikia.com/wiki/Loader%27s_number
[2] https://github.com/rcls/busy