IMHO an output of a function being bigger than Graham's number isn't particularly impressive.

Consider TREE(3) -> https://www.youtube.com/watch?v=3P6DWAwwViU

Graham's Number is notable for being one of the first, being an actual relevant bound in an unrelated proof (not just the output of a function), of teaching people the Up Arrow Notation.

To me, BB(10), TREE(3) etc. being so large is only minorly more impressive than GG(1) where I define I define GG(x) as "Graham's Number + X"

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guipsp 6 hours ago | parent | next [-]

That is not the reason it is interesting. Rather, it is interesting because it (at least claims) a new bound on BB(10)

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Jaxan 5 hours ago | parent [-]

I don’t think such a bound in itself is interesting, unless the proof reveals some strategy to prove more. There are infinitely potential bounds to BB(n), why is this particular case BB(10) > G useful for understanding fast growing functions?

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Scarblac 4 hours ago | parent | next [-]

It's impressive that a Turing machine of only 10 states can run for Graham's number steps and still halt.

I'd call it absolutely mind boggling, actually.

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NooneAtAll3 5 hours ago | parent | prev [-]

> why is this particular case BB(10) < G useful for understanding fast growing functions?

it's... not?

it is useful for understanding BB()

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Jaxan 5 hours ago | parent [-]

First, an upper bound would be more useful to understand BB. But second, out of all large numbers, does this one give any information? We’ll have to see. The fact in itself is useless.

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NooneAtAll3 4 hours ago | parent [-]

> First, an upper bound would be more useful to understand BB.

you can't have upper bound without breaking Turing and solving halting problem

> But second, out of all large numbers, does this one give any information? The fact in itself is useless.

yes, it gives info of "OMG BIG NUMBER I KNOW ABOUT, THAT'S SO COOL" quality, like all the rest of recreational mathematics :)

Graham's number is simply a nice checkpoint that invites everyone to learn more about function hierarchies

	▲	Jaxan 4 hours ago \| parent [-]
		> you can't have upper bound without breaking Turing and solving halting problem Not in general, no. But for a fixed n, you can. For example, people have settled the exact value of BB(5).

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raincole 6 hours ago | parent | prev [-]

TREE(3) is interesting as it's not very intuitive that it's even a finite number. Graham's Number is obviously finite.