Unfortunately I don't understand that Wiki page.
The page links to https://en.wikipedia.org/wiki/Friedman%27s_SSCG_function , which is supposedly even bigger but I can't follow the text of the first page so won't attempt to read that.
Unfortunately I don't understand that Wiki page.
The page links to https://en.wikipedia.org/wiki/Friedman%27s_SSCG_function , which is supposedly even bigger but I can't follow the text of the first page so won't attempt to read that.
His other channels are worth checking out too, Sixty Symbols and Deep Sky Videos being my favorites. To be clear, I have no affiliation with this, just a fan.
[1] https://www.youtube.com/watch?v=HX8bihEe3nA&list=PLt5AfwLFPx...
I especially like the intro to Graham's number and the rough proof that it is unthinkably large, that your head would literally implode into a black hole an uncountable amount of times over.
Math is so boggling. Numbers that are literally super-natural exist and are 'useful' at the same time, yet are impossible to mortals all the same. It's this weird end-around we manage with math/logic. That 'floor falling out from underneath you' feeling, a happy giddiness and smallness.
Sorry, Graham's Number does something fierce to my head, it seems.
It also has the geek cred of being represented by an obfuscated C program (the unobfuscated verson is also available[2]).
k(1) = TREE(1), K(2) = TREE(TREE(2)), K(3) = TREE(TREE( <(repeats (TREE(3) time (TREE(3)))
Sequence 1, 2, non comprehensibly large number.
Now define m(x) in terms of k and m(x-1).
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.
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.
So, they are all going to be take large value and use it to boost some other function m(x). I just happen to pick a large f(x).
BTW, I totally geeked out reading about Graham's Number. Thanks to the OP!
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.
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]
https://blog.regehr.org/archives/140
C Compilers Disprove Fermat’s Last Theorem
> SSCG(3) is not only larger than TREE(3), it is much, much larger than TREE(TREE(…TREE(3)…)) where the total nesting depth of the formula is TREE(3) levels of the TREE function.
notwithstanding that TREE(3) is pretty crazy in itself:
> an extremely weak lower bound for TREE(3), is A(A(...A(1)...)), where the number of A's is A(187196), and A() is a version of Ackermann's function. Graham's number, for example, is approximately A^64(4)