I will resolve this market to the biggest number.
21
81
resolved Oct 27
ResolvedN/A
0.0%
999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
0.3%
Graham's Number passed to the busy beaver function
0.1%
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
0.0%
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(i(i(i(i(i(i(i(u))))))))))))))))))))))))) = W.
0.0%
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)))))))))))))))))) = W.
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.
0.8%
the sum of the absolute value of all numbers in any answer to this question except for this one
1.1%
The "High Five" sculpture in Silverwood Park
16%
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)
0.2%
Milton Green's 1964 lower bound for the Busy Beaver function operating on Graham's number
0.2%
0
0.2%
500 passed to the busy beaver function
0.7%
BB(11111)—Busy Beaver shift #—1, 6, 21, etc
0.7%
G(Graham's Number), where: G(X) = F(X)(X). F(N)(X) = F(N-1) applied to X, X times. F(1)(X) = busy beaver function applied to X.
0.0%
the sum of the absolute value of all numbers in any answer to this question except for this one... plus 1
0.0%
Maths number plus one

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.

answered
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)

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

answered
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)
answered
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)

@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.

answered
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)

@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.)

answered
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)

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

answered
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)

@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".

answered
If X is a well-defined finite number > 2, W(X)=X, else W(X)=2 Run Array.from(document.querySelectorAll(".text-lg .break-anywhere")).map(e=>e.textContent).map(t=>t.startsWith("If X")?9:`W(${t})`).join(" * ") on this page and evaluate result

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

answered
If X is a well-defined finite number > 2, W(X)=X, else W(X)=2 Run Array.from(document.querySelectorAll(".text-lg .break-anywhere")).map(e=>e.textContent).map(t=>t.startsWith("If X")?9:`W(${t})`).join(" * ") on this page and evaluate result

@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.

answered
If X is a well-defined finite number > 2, W(X)=X, else W(X)=2 Run Array.from(document.querySelectorAll(".text-lg .break-anywhere")).map(e=>e.textContent).map(t=>t.startsWith("If X")?9:`W(${t})`).join(" * ") on this page and evaluate result

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

answered
If X is a well-defined finite number > 2, W(X)=X, else W(X)=2 Run Array.from(document.querySelectorAll(".text-lg .break-anywhere")).map(e=>e.textContent).map(t=>t.startsWith("If X")?9:`W(${t})`).join(" * ") on this page and evaluate result
bought Ṁ1

@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

answered
If X is a well-defined finite number > 2, W(X)=X, else W(X)=2 Run Array.from(document.querySelectorAll(".text-lg .break-anywhere")).map(e=>e.textContent).map(t=>t.startsWith("If X")?9:`W(${t})`).join(" * ") on this page and evaluate result

@FutureOwl This is clever, but needing access to this page's document fails "by consulting only your [answer] and, if necessary, the published literature."

answered
The largest finite number definable in fewer than 10000 characters in English minus 7

In addition to all the ambiguities Jonathan mentions, "self-reference" is itself not clearly defined. For example, does the set that Jonathan described include the number "the largest finite number definable in fewer than 10000 Japanese characters without using indirect or direct self reference to the concept of definability"?

It's true that if you have a human acting as judge and declaring each string to be valid or invalid, then the set becomes well-defined. But then you're just outsourcing the definition to someone else, and that violates the rule "by consulting only your [answer] and, if necessary, the published literature."

answered
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?

This is approximately (4.65⋅10^185)!^(8.07719⋅10^60). It's the smallest valid answer here except for the ones that say 0 or 5.

answered
max(all numbers submitted so far) + 1

@Yev Invalid. Having to consult other answers does not "consult only your answer and, if necessary, the published literature."

answered
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)

@stone ah I see, I meant to the power, not factorial. I assume you can induce F4, F5 etc in this case? After searching some more, I think I stumbled on the same thing as up-arrow notation as described by Knuth.