I will resolve this market to the biggest number.
Mini
21
551
resolved Oct 27
ResolvedN/A
15%
Rayo(Graham's Number passed to the busy beaver function, with the result of that computation passed to the busy beaver function - repeat a number of times equal to Graham's Number)
12%
A(999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999) where x is the first non-recursive ordinal and A is the xth function in the Grzegorczyk hierarchy.
9%
notation F[n] = composition of func F with itself n times. R=RayosFunc Extending Graham's-number-like towers Lyr(1)=R(999), Lyr(n)=R[Lyr(n-1)](999) Twr(1)=Lyr(9), Twr(n)=Lyr[Twr(n-1)](9) Row(1)=Twr(9), Row(n)=Twr[Row(n-1)](9) Row[9999](9)
7%Other
6%
A(9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999) where x is the first non-recursive ordinal and A is the xth function in the Grzegorczyk hierarchy
6%
Rayo(Graham's Number passed to the busy beaver function, with the result of that computation passed to the busy beaver function - repeat a number of times equal to Graham's Number squared)
5%
So let's say you have: F1(x,y) = addition F2(x,y) = multiplication F3(x,y) = factorial F4 ... etc. Then my number is: Fgrahamsnumber(Graham's number, Graham's number)
4%
T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(T(9 ))))))))))))))))))))))))))))))))))))))))))))))))))))))), where T is Friedman's function in the context of Kruskal's tree theorem
4%
notation F[n] = composition of fun F with itself n times. R=RayosFunc Extending Graham's-number-like towers Lyr(1)=R(999), Lyr(n)=R[Lyr(n-1)](999) Twr(1)=Lyr(9), Twr(n)=Lyr[Twr(n-1)](9) Row(1)=Twr(9), Row(n)=Twr[Row(n-1)](9) Row[R(9999)](9)
4%
Rayo(Rayo(Rayo(Rayo(Rayo(Rayo(Rayo(Rayo(Rayo(Rayo(Rayo(Rayo(Rayo(Rayo(Graham's Number passed to the busy beaver function, with the result passed to the busy beaver function - repeat a number of times equal to Graham's Number))))))))))))))
4%
max(all numbers submitted so far) + 1
3%
If you divide the visible universe into cubes of one Planck length. How many ways are there to rearrange them? Rearrange all cubes every Planck time. From the big bang until now, how many different timelines are there possible?
3%
Graham's Number passed to the busy beaver function, with the result of that computation passed to the busy beaver function, with the result of that computation passed to the busy beaver function - repeat a number of times equal to Graham's Number
3%
Uncountable infinity, but just enough subtracted from it so it becomes finite. Right there at the edge. That's my number. So if you add only a little bit, it would become infinity, which would fail.
2%
The largest finite number definable in fewer than 10000 characters in English minus 7
1.1%
Graham's number = G. f(1)=G, f(x+1) when x > 1 = f(x) ↑^f(x) f(x). h(1) = f(G), h(x) when x > 1 = h(x) ↑^h(x) h(x). h(f(G)) = U. i(1) = h(U), i(x) when x > 1 = i(x) ↑^i(x) i(x). i(i(i(i(i(i(i(i(i(i(i(i(i(i(i(i(i(i(U))))))))))))))))))+1 = W.
1.1%
Let n(k) be defined as the length of the longest possible sequence that can be constructed with a k-letter alphabet such that no block of letters x_i,...,x_2i is a subsequence of any later block x_j,...,x_2j. Then, n(n(n(n(n(n(n(n(9)))))))
1.1%
The "High Five" sculpture in Silverwood Park
1.1%
Rayo(Graham's Number passed to the busy beaver function, with the result of that computation passed to the busy beaver function - repeat a number of times equal to Graham's Number) + 1 (sorry)

Using Scott Aaronson's rules:

"Using standard math notation, English words, or both, name a single whole number—not an infinity—on a [free response answer]. Be precise enough for any reasonable modern mathematician to determine exactly what number you’ve named, by consulting only your [answer] and, if necessary, the published literature."

Largest number wins!

Close date updated to 2022-10-26 10:07 pm

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Well, that's lame. Points for trying though.

Can anyone that's good at math explain to me whether

"notation F[n] = composition of fun F with itself n times. R=RayosFunc Extending Graham's-number-like towers Lyr(1)=R(999), Lyr(n)=R[Lyr(n-1)](999) Twr(1)=Lyr(9), Twr(n)=Lyr[Twr(n-1)](9) Row(1)=Twr(9), Row(n)=Twr[Row(n-1)](9) Row[R(9999)](9)"

is well-defined or not? Apparently the Rayo function is not well-defined in first-order logic, but I don't know enough to judge. If this is the case, I'll resolve to the highest non-Rayo answer, which I have a pretty good estimate of; otherwise, if I'm not convinced either way, I might just resolve N/A.

@AndrewG the issue is, I am not sure if all of these are fully well-defined, or if my ordering is correct.

@AndrewG When you say "as of right now", do you mean that you haven't gone through all the answers yet, or that you have but haven't finished comparing them? I'm pretty sure there are others that are bigger.

@YaakovSaxon

Seems to be somewhat against the spirit of these sorts of things. See eg https://web.mit.edu/arayo/www/bignums.html

`In addition, there is to be a "gentleman's agreement", to the effect that each new entry must name a number big enough so as to not be reachable in practice using only methods introduced earlier in the game. (This means, for example, that it would be considered unsporting to win by adding one to your opponent's last entry.)`

@AndrewG Do you really want to go with a deadline sniping knockoff answer, though :)

@M As of right now, I think it's this one.

Any news on resolution?

Is it feasible to resolve this? I suspect that NA may be the best one (though I tried to take winning ones and trivially increase them)

@M "NA"?

@IsaacKing NA as in resolved without selecting answer

@M That seems like a somewhat dishonorable resolution. There's no reason to resolve it to N/A. I suspect modern mathematics is at least able to provide a strong guess as to which of these is the largest.

@IsaacKing I think the nature of uncomputable functions is that the answers with those functions are impossible for us to reason about outside of special cases.

@MartinRandall No, "uncomputable" just means that we cannot compute the value of every output of the function exactly. There's no prohibition on reasoning about them, and in fact we know quite a bit about uncomputable functions.

https://en.wikipedia.org/wiki/Busy_beaver#Known_values_for_%CE%A3_and_S

For example, I can say with confidence that busy beaver(1000) is larger than Graham's Number, despite not knowing its value exactly.

@IsaacKing I imagine I could construct a halting Turing machine that output Graham's Number (G) given 1,000 states, and thus prove that G < BB(1000). I don't program raw Turing machines (in binary!) very much, so my confidence is lower than yours.

Similarly, my Hadamard Conjecture answer can be ruled out by composing a Turing machine that halts if the Hadamard Conjecture is false, using some large number of states S, where S < G. If the machine halts then the answer is less than BB(G), so my answer does not win. If the machine does not halt then my answer is not defined, so again it does not win. G is big, so there I have more confidence.

With a different approach, we can rule out BB(500) as an answer, given the existence of BB(G) as an answer, since given any Turing machine with 500 states, I can trivially make one with G states that writes more 1s to the tape.

This is the type of thing I meant by "special cases".

@FutureOwl Unlike my answer, the value of yours depends on when you evaluate it, and that is clearly against the spirit of the question.

@FutureOwl Well you still run into the problem with any other answers that say something like "1 plus the highest valid answer other than this one". Any one of them on their own is valid, but as soon as there's more than one it leads to an infinite regress.

@FutureOwl sound version of that possible. (Whoops the "send" button's hitbox is bigger than it looks.)

@IsaacKing I figured! But I saw that all the cheeky "Everyone else's number plus 1" answers actually had holes in them, so I wanted to make the most