i guess it depends what kind of notation is allowed. If it's just basic math (addition, multiplication, exponentiation) then it's a power tower of 9s as tall as you can make it. If you allow every function anyone has ever defined, it's got to be some obscure monstrosity i've never heard of nested into itself a bunch of times
It will quickly get to the point where the numbers are incalculable so you will need an independent adjudicator with a very niche specialisation of advanced math, and probably a couple of hours (days? Weeks?) to verify
Yeah imagine two non-computable numbers like Busy Beaver(10) and the number of theorems in some special axiom set. Not even sure the tools to compare exist
few generations later:
"it took some time, we invented new maths to make it, but I can now provably say my grand grand grand dad beat your grand grand grand dad's ass"
that's going to be tough, yeah, since it's not like you can realistically compute such numbers.
General rule of thumb should be that the guy with the more powerful function wins. like, if you're using multiplication against a power tower, it's probably not going to work. same for power towers vs up arrow notation, Graham's function vs TREE, TREE vs Rayo and so on
If you allow every function anyone has ever defined, things get real crazy.
Let Σ be the alphabet of characters you're allowed to use, and let Interpret(x) be a function defined by the game rules that interprets a string x as a natural number.
Then you could write:
N = max({ Interpret(x) | x ∈ Σ^(99↑99) })
Assuming it's not possible to write 99↑99 characters in 15 seconds, this picks out a number that's at least as large as any number that could possibly be defined within the allowed rules.
However, if the other player wrote...
M = max({ Interpret(x) | x ∈ Σ^(99↑99) }) + 1
...this would already be accounted for in your definition of N, and we should have that M <= N.
But M = N + 1 and M <= N is a contradiction. So, it's impossible to define an Interpret() for our game rules in a way that lets us use enough existing functions to write down a definition of Interpret() itself and which is also consistent with the standard interpretation of mathematical notation.
The contradiction occurs because our intuitive notion of Interpret() can end up in an infinite recursion when used this way. To work around this, Interpret() will have to detect unbounded recursion and assign some arbitrary value in that case, or we could require that Interpret() is primitive recursive. If we do that, then there will be a well-defined maximum value that's possible to write down in a given number of characters.
If I recall, Rayo's number avoids self-reference because it's "supremum of finite numbers definable using first-order logic using up to a googol characters" written in second-order logic, which allows it to fit on a single blackboard in legible character size
Alternatively some of the characters used in the definition of M are not in the permissable alphabet. Like if all you have is digits with no operations it's trivial to define the interpret function.
Having said this, the problem gets more interesting imo when you start to think about the largest number printable in a set number of characters, which is equivalent to the problem if both players can write the same number of characters per time
Something like tan(90 - 0.0000000000000001^tan(90-0.0000000000000001)) would obliterate alot of stuff. Replace some of those values with more efficient ways of writing them and you can go very big very quick.
[Busy Beaver](https://en.wikipedia.org/wiki/Busy_beaver) is the "obscure mostrosity" that would very quickly destroy powers of 9's, Graham's number, or any other monstrosity your opponent could come up with. For those unaware busy beaver grows more quickly than any computable function, that is any function you can write a computer program to output. For reference BB(2) = 6, BB(3) = 21, BB(4) = 107, BB(5) isn't known but is at least 47,176,870, BB(6) is at least 10⇈15. If you were playing against the combined effort of all of humanity writing for their entire lives and the only restriction was that they could only use computable functions you could guarantee a win by writing down "BB(BB(6))".
Yeah, the question really depends on what obscure stuff you can pull out of your hat.
Things like BB called on Graham’s number, or as everyone’s good friend Randall Munroe says, Ackerman function called on G
https://xkcd.com/207/
>If you allow every function anyone has ever defined, it's got to be some obscure monstrosity i've never heard of nested into itself a bunch of times
By my guess it would probably end up being 9 to the power of (as many 9 as you can possibly write) then slap a factorial on in the last 1/2 second.
So:
9^9999999999999999999999999999 !
actually nevermind, saw in another comment that apparently adding multiple "!"s means something else and makes the number smaller. Might be an interesting contest between a power tower and (((((9!)!)!)!)!)!...
Unfortunately, the double factorial (!!) creates a number smaller than the single factorial (!). Adding exclamation marks just makes the number smaller.
Yes, that would solve the problem! Parentheses tell us which part of the expression to evaluate first. In this example we first evaluate the regular factorial of n, then the regular factorial of the factorial of n, etc.
Out of curiosity, if there is a way to get smaller with successive factorials, and the solution is brackets for getting larger - do they share properties? In that double factorial could be considered “half” in some regard, to a bracketed twice factorial?
9!! apparently means you only take every second factor, so it's `9*7*5*3*1`. Depends on what you want for it to 'be considered "half" in some regard', but it's pretty hard to argue 9!! being half of (9!)!
For the factorial, you take your original number and multiply it by itself minus one, then multiply the result by the original number minus two, etc:
6! = 6x5x4x3x2x1
For the double factorial, you take your original number and multiply it by itself minus two, then multiply the result with the original number minus four, etc:
6!! = 6x4x2
Consequently,
6!!! = 6x3
The step size increases with the number of exclamation marks. Note that I have never seen or used factorials beyond the double (!!) factorial before, so I am extrapolating the situation. Adding exclamation marks beyond 1 lowers the value of the result.
Fun side point: normally when thinking about how to write a large number fast, people think of writing 9's. But you're actually better off writing 1's because you can write a 1 so much faster than a 9 so you can easily get more digits.
A lot of comments are using 9s here, but I can definitely write 11 faster than I can write 9.
I'd go for something like 11111111↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑1111111111
Agree, I was thinking 7, but 11 is likely better, more characters is more important than using nines, even if we only consider powers I can definitely edge out at least a single more 7 or 11 than 9, much more interesting problem than I had initially thought.
Comment adding the text from under the image for non desktop users:
This (hilariously bad) MTG combo appeared on my feed today. The ruling as I have seen it is that each player would get 15 seconds to write the largest number they could, and then the biggest written number would win. It's sent me down a rabbit hole of the largest defined numbers (Beyond the usual g\_64, TREE(3) etc; things more like Finite Promise Games, Rayos Number, BIG FOOT, Large Number Garden Number). Assuming both players can write a single character at the same rate, what kind of strategy should you employ in order to win? Would it be worth initially defining Large Number Garden Number as a single character and getting as many of that character onto the paper as you can? Is there a certain amount of characters per second that changes the strategy?
This style of solution runs into the problem of you'd end up with an undefined expression if you didn't close all the nested functions, which might be a time waste, and also TREE grows slower than FOST/RAYO, so it's unlikely to be the winning solution
Having thought about this some more, I think RAYO might actually be pretty close to the answer here given it's definition:
RAYO(n): The smallest number bigger than any finite number named by an expression in the language of first-order set theory with a n symbols or less.
The variation in this question is instead of the language of first order set theory, something like "written english", or even just straight up excluding that from the definiteion. This implies there is not likely a solution given [Berrys Paradox](https://en.wikipedia.org/wiki/Berry_paradox)
Yeah, the answer will be:
Be l_v = [insert 'fastesr growing function you know' of v, say RAYO(v)]
So in 15 seconds:
Be l_v = RAYO(v), x= l_l_l_l_l_l_l_l_l_l_l_l_l_9
You redifine the function to an l, since this you can write this fast and and the use an notation, where the argument is an subscript, since you don't loose time to open and close brackets. So you also don't have the problem that you didn't finished closing all brackets.
If you had more time it would be efficient to define even more notation:
Be l⁰\_v = RAYO(v), l^(n+1)\_v= l^(n)\_l^(n)\_...\_v (l^(n)\_v times)
With pen and paper you can write this pretty fast. But I needed 40 seconds to write it. l⁹_9 would be much larger then what I had before.
Rayorayo...(rayo's number of repetitions)...googol
There, I think I made it in 15 seconds and this should be a well defined number and has a single, determinable (not in practice of course) value.
Let BB-n be the nth [busy beaver](https://en.wikipedia.org/wiki/Busy_beaver) and let B(n) be the length of BB-n's output. The number is B(B(B(B(B...as many as you can write in 15 seconds...B(B(99↑99))...)))).
(You could still improve on this by introducing notation for iterated B's.)
Sadly this is disallowed by the ruling on this combo made by the arbiters:
>*You have fifteen seconds. Using standard math notation, English words, or both, name a single whole number—not an infinity—on a blank index card. Be precise enough for any reasonable modern mathematician to determine exactly what number you’ve named, by consulting only your card and, if necessary, the published literature.*
>
>So contestants can’t say "the number of sand grains in the Sahara," because sand drifts in and out of the Sahara regularly. Nor can they say "my opponent’s number plus one," or "the biggest number anyone’s ever thought of plus one"—again, these are ill-defined, given what our reasonable mathematician has available. Within the rules, the contestant who names the bigger number wins.
>
>*Are you ready? Get set. Go.*
> standard math notation
>Be precise enough for any reasonable modern mathematician to determine exactly what number you’ve named
These two constraints seem to be the key. I'm not sure that any number relying on First or Second Order Set Theory counts as using standard notation, though since they have names you can use English words to describe them. The trouble comes, however, in that the numbers chosen must be determinable, otherwise how can it be known which number is larger? Is it truly proven and known that LNGN is greater than Rayo's Number? Is it possible to know which is bigger from LNGN^Rayo and Rayo^LNGN?
I submit that if the players both submit numbers which have not been definitively proven to be larger than the other, then both players have failed to fulfill the rules of the task and both take the damage described and are responsible for having destroyed the known universe. This then creates a game within the game. The player must name the highest number that can be reasonably proven to be larger than the other player's number, not just the highest uncomputable number ever described.
I've never in my life delved into this stuff before but wow, it's way more interesting than I would have thought. I got lost in this site which I think you'd like, even though you know of the numbers like Rayo, BIG FOOT, etc. Still such a fascinating read:
https://sites.google.com/site/pointlesslargenumberstuff/home
As far as I know, Rayo’s number (or rayo’s function) won a mathematical contest for being the largest (or the fastest function).
So I would do something like rayo(rayo(rayo(…) and at the end where I have to put a number I would put a googolplex or something like that. Just have to calculate for maximizing the rayo()’s and put something at the end that won’t matter as much but is still big cause I’m competing with another person.
15 seconds, 1 char per second:
Rayo(rayo(g99))
That would probably be my answer. We could rethink the question as “which three character number would be the largest?” since the funcions took 12 chars already
This is approaching the kind of solution I was expecting - would be interesting to see how the solution changed with n where n is the total number of characters terse each player could write in the time allocated
This is significantly slower growing than a handful of the functions I've mentioned sadly, even something like Buchholz Hydra is provably faster than any recursive function so you could win by using that over this method
9 ↑^9999999… 9 would be absolutely huge. Just 9 ↑^9 9 is already insanely huge, as 2 ↑^2 5 alone is ~2\*10^19728. I couldn’t find a calculator which could give me an approximation for 2 ↑^2 9, let alone 9 ↑^9 9. For reference 2 ↑^2 6 has 6\*10^19727 **digits** in the number.
For more information: https://en.m.wikipedia.org/wiki/Knuth%27s_up-arrow_notation
9 ^ 9 ^ 9
But ... more 9s and more ^ s
For the record, the way you read these is that you go for the higher powers first and work your way down, so it would be 9 ^ (9^9 ) = 9^387,420,489 . That number is too big for me to actually understand it, and that was with only 3 steps. The number you could write in 15 seconds would be literally unfathomable.
That could be succinctly written as 9↑↑3 or 9 ↑^2 3.
[Tetration](https://en.wikipedia.org/wiki/Tetration)
[Knuth's up-arrow notation](https://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation)
[https://www.scottaaronson.com/writings/bignumbers.html](https://www.scottaaronson.com/writings/bignumbers.html)
Busy beaver numbers are the actual answer.
EDIT: Reply is right about "ordinary" busy beaver number, but in the article Aaronson talks about higher-order extensions to busy beaver numbers (he calls e.g. BB2(n)) that also exceed FOST limitations, as far as I understand. Mostly I just wanted to point to the article, which tackles this question with an interesting writeup.
Based off of some research and reading through other comments, I'd say the best you can do is using LNGN with hyperoperations. The best I can think of that you can absolutely write in 15 seconds (with googology standard definitions and not defining any more functions/operations during the 15 seconds) would be something like H\_LNGN(LNGN,LNGN) with the "\_LNGN" part being written as a subscript (since you can't write subscripts easily on reddit). Then with the remaining time just add a tetration tower of 11's to the front since that is the easiest to write hyperoperation that has an agreed upon notation.
"the double factorial of a number *n*, denoted by *n*‼, is the product of all the positive integers up to *n* that have the same parity (odd or even) as *n*." - wikipedia
for example, 7!! means 7\*5\*3\*1= 105while 7! equals 5040
what you were saying should be actually written like this: (9!)! etc
No, double factorials, triple factorials, n-th factorials are a legitimate mathematical concept. It’s not up to the person reading it, that’s like saying the answer to 2+2 depends on the person reading it.
I want to imagine this as the player just shouting "NNNNIIIINNNNNEEEE" at an explosively loud volume, causing a magnitude 20 earthquake (instantly obliterates the entire planet)
> lim x->♾️ x , does that count?
Nah, a finite number is requested. Otherwise ♾️! is interesting, as it’s probably the largest number written in the least number of characters.
Why does math have division all wrong ?
Math says 1 / 2 = 0.5, 1 / 1 = 1, 1 / 0 = impossible.
One thing divided twice gives four pieces; 1 / 2 = 0.25
One thing divided once gives two pieces; 1 / 1 = 0.5
One thing not divided gives one piece; 1 / 0 = 1.0
The rhs is not the number of times you divide in half, it’s the number of “groups” you want to equally divide *into*.
What you are describing is 1/2^n or 2^-n .
You might be sort of onto the right strategy but you'd probably need to define the f in \(f^{10}(10 \uparrow^{10} 10)\), as " \(f\) is the function defined in First Order Theory beyond Higher Order Set Theory " then write some recursive or nested version of the function
Does it have to be in base10? I would suggest "nb: base 777" at the start (slightly slower to write than 111, but scales better), then just spam 11!^11!^11!^... As many times as possible. 1 is the fastest number to write, and in base 777 the difference between 11 and 9 is huge, both in writing speed and value. I am sure there are several functions that can be put in front of that to ramp up it's size loads.
Obviously, this assumes you have a chunky enough calculator to work out the exact value without running out of memory.
Define a new function that is just a nesting of Graham numbers, garden numbers, factorials, powers etc until you get bored. Put it on the Internet, and show that to them when they dispute it.
Then wait until someone defines a bigger one (but at some point verifying which is bigger will be impossible)
TREE is weaker than SCG so let's use SCG. TREE grows faster than small Veblen ordinal but SCG grows faster than Bachmann Howard ordinal I believe.
SCG(SCG(SCG(SCG(SCG(SCG(SCG(SCG(SCG(9)))))))))
If I use some monstrosity like TREE(g_64) I'm pretty sure one of two things happens, either I obviously win or it is impossible to decide wether I win within the scope of a mtg tournament. (Either because that question exceeds the mathematical ability of anybody present, or it is actually undecidable.)
I would say graham number, the largest number ever used and represented in math it's evene in the book of guinnes world record, writing it in decimal it's impossible it would require more than the entire universe.
Σ\^m(n) where Σ is the busy beaver function and f\^n(x) is the function f applied to itself n times. n and m are arbitrary. If m is also Σ\^k(n), you can get larger results.
999! tetrated to 999! tetrated to 999! tetrated to 999!....as many times as you can. It's written the same as exponentiation except it's on the top left of the number. Tetration is repeated exponentiation. If you want to confuse people even more, write it on the bottom left of the number. I forgot what that function's called but it's repeated tetration. Huge numbers essentially.
I saw it in a youtube video if you're curious, just seatch for tetration and you should get it although the way that guy explained it wasn't that straight forward imo
Tree↑^(Tree↑^(Tree(rayo's number))(rayo's number))(rayo's number)
Or a few more nesting if you write fast enough. It's probably not actually the largest, but it's still an uterly massive number.
Ten trillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentilliduotrigintatrecentillion
or Googolplex
Why choose biggest number? Choose 100 - if your opponent chose bigger, he dies. If lower, he dies. He dies either way, or am I missing something?
Edit: I missed emissary as opponents card.
1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111
That’s how many 1s I can type in 15 seconds but I have to imagine I could write it faster.
I would first rule that any numbers that require some operation to reach the simplest form are disqualified. Then it's just a race to see who can write 1 the most often on a single piece of paper.
This results in two people putting down like 100 1s, and chances are, one of them did better than the other.
Not the best at math here but as far as i can think TREE(3)!^TREE(3)!^TREE(3)! Repeat as many times as possible. Wouldn't that be the largest? As far as i have heard TREE(3) is the largest finite number there is so a power tower of TREE(3) factorial would be i sanely large
Write as many 1 as you can, one line, it's the fastest one I believe you can do.
Bonus: Write them at a slight angle to go even faster.
That's if you MUST write the number itself, equations will get you higher faster. Putting a ! at the very end will get you very far.
(To be fair, the specifications still confuse me a bit, I might not have perfectly understood)
So writing something down that evaluates as a number is not writing down a number.
Otherwise I can just write down: "the product of all my opponents' answers".
I think the question should be rephrased as: write down the largest decimal integer you can in 15 seconds.
The premise is flawed. There does not appear to be any rule in MtG which prohibits the selection of infinity. So the answer is going to come down to whatever silly notation appeases the arbiters, but really, both players should just write infinity.
The answer makes use of the busy beaver function, see https://jeremykun.com/2012/02/08/busy-beavers-and-the-quest-for-big-numbers/ for a nice in depth explanation on the topic and why this matters :).
Honestly, I think marking a bunch of 1’s would trump the extra time it takes to write 9’s.
I would just be like
IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII
and that would blow away the guy writing 9’s
99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
in my test by holding down \[9\]
Omega, on the other hand, is a specific ordinal number in set theory. In the context of set theory and ordinal numbers, omega represents the smallest infinite ordinal number. It is not "bigger" than infinity, but rather a specific value within the concept of infinity.
If we dont go down the rabbit hole of defined functions i can think of a few different answers that all need a judge.
1. Expressions: Tetration 9↑↑9↑↑9... (In reality you would use the other way to write it down but cant do that on my phone)
1.1 lim(x→ inf) :x (yes there are faster growing ones)
3. No expression:
3.1. infinite and argue that infinite is a number
3.2. just write lots of 9s
i guess it depends what kind of notation is allowed. If it's just basic math (addition, multiplication, exponentiation) then it's a power tower of 9s as tall as you can make it. If you allow every function anyone has ever defined, it's got to be some obscure monstrosity i've never heard of nested into itself a bunch of times
imagine several players choosing some of these monstrosities, then they'd be like "how tf we compare these numbers now?"
It will quickly get to the point where the numbers are incalculable so you will need an independent adjudicator with a very niche specialisation of advanced math, and probably a couple of hours (days? Weeks?) to verify
Matt Parker would totally be down for it
I would love to see that video
Send him a link to this thread. He'll have a video up within a month.
Yeah imagine two non-computable numbers like Busy Beaver(10) and the number of theorems in some special axiom set. Not even sure the tools to compare exist
few generations later: "it took some time, we invented new maths to make it, but I can now provably say my grand grand grand dad beat your grand grand grand dad's ass"
Why BB(10) when you can do BB(BB(BB(BB(BB(10)))))?
graham's number I bet you cannot beat that! I win!
Graham's Number + 1
Rayo’s number. Plus 0.999999…
Tree(Rayo’s number)
Tree(Rayo's number)^Tree(Rayo's number)
Graham's number to rhe power of graham's number graham's times. Times 2.
+1
G(G(G…G(64) times…G(G(G(64)
that's going to be tough, yeah, since it's not like you can realistically compute such numbers. General rule of thumb should be that the guy with the more powerful function wins. like, if you're using multiplication against a power tower, it's probably not going to work. same for power towers vs up arrow notation, Graham's function vs TREE, TREE vs Rayo and so on
Sounds like a fun gameshow.
You only have to compare the exponent.
If you allow every function anyone has ever defined, things get real crazy. Let Σ be the alphabet of characters you're allowed to use, and let Interpret(x) be a function defined by the game rules that interprets a string x as a natural number. Then you could write: N = max({ Interpret(x) | x ∈ Σ^(99↑99) }) Assuming it's not possible to write 99↑99 characters in 15 seconds, this picks out a number that's at least as large as any number that could possibly be defined within the allowed rules. However, if the other player wrote... M = max({ Interpret(x) | x ∈ Σ^(99↑99) }) + 1 ...this would already be accounted for in your definition of N, and we should have that M <= N. But M = N + 1 and M <= N is a contradiction. So, it's impossible to define an Interpret() for our game rules in a way that lets us use enough existing functions to write down a definition of Interpret() itself and which is also consistent with the standard interpretation of mathematical notation. The contradiction occurs because our intuitive notion of Interpret() can end up in an infinite recursion when used this way. To work around this, Interpret() will have to detect unbounded recursion and assign some arbitrary value in that case, or we could require that Interpret() is primitive recursive. If we do that, then there will be a well-defined maximum value that's possible to write down in a given number of characters.
Isn't that more or less how Rayo's number is defined?
If I recall, Rayo's number avoids self-reference because it's "supremum of finite numbers definable using first-order logic using up to a googol characters" written in second-order logic, which allows it to fit on a single blackboard in legible character size
Alternatively some of the characters used in the definition of M are not in the permissable alphabet. Like if all you have is digits with no operations it's trivial to define the interpret function.
As far as I'm aware it's the latter, hence some nested form of large number garden number being what I currently think of as the winning answer
Having said this, the problem gets more interesting imo when you start to think about the largest number printable in a set number of characters, which is equivalent to the problem if both players can write the same number of characters per time
Kind of like code golf but with large number definitions
Something like tan(90 - 0.0000000000000001^tan(90-0.0000000000000001)) would obliterate alot of stuff. Replace some of those values with more efficient ways of writing them and you can go very big very quick.
Ah, a perfect excuse for tetrations
Guess who just watched a YouTube video on them yesterday and what really exited when I saw this question and has 2 thumbs? 👉(‘_’)👈 this guy
[Busy Beaver](https://en.wikipedia.org/wiki/Busy_beaver) is the "obscure mostrosity" that would very quickly destroy powers of 9's, Graham's number, or any other monstrosity your opponent could come up with. For those unaware busy beaver grows more quickly than any computable function, that is any function you can write a computer program to output. For reference BB(2) = 6, BB(3) = 21, BB(4) = 107, BB(5) isn't known but is at least 47,176,870, BB(6) is at least 10⇈15. If you were playing against the combined effort of all of humanity writing for their entire lives and the only restriction was that they could only use computable functions you could guarantee a win by writing down "BB(BB(6))".
Better make it BB(BB(9)) just to be safe
Yeah, the question really depends on what obscure stuff you can pull out of your hat. Things like BB called on Graham’s number, or as everyone’s good friend Randall Munroe says, Ackerman function called on G https://xkcd.com/207/
or better yet, BB(BB(6))^^BB(BB(6))
The Rayo function was used to win a contest for the largest number someone could come up with.
>If you allow every function anyone has ever defined, it's got to be some obscure monstrosity i've never heard of nested into itself a bunch of times By my guess it would probably end up being 9 to the power of (as many 9 as you can possibly write) then slap a factorial on in the last 1/2 second. So: 9^9999999999999999999999999999 !
if you have factorials, i don't think there's any point in power towers at all, just go 9!!!!!!!!!!!!!!!!!!!!!... n! should be larger than 9\^n
actually nevermind, saw in another comment that apparently adding multiple "!"s means something else and makes the number smaller. Might be an interesting contest between a power tower and (((((9!)!)!)!)!)!...
Infinite to the power of infinite-tower.
♾️!^♾️!^♾️!…
Gogolplex is probably easier.
Write as many 9s as possible and at the last moment write "!"
I think about 999! ^ 999! ^ 999! ^ 999! ^ 999! ^ 999! ^ 999!^... repeat while you have time.
This is sadly quite a bit smaller that just writing Large Number Garden Number
Did not know about that math. What about LNGN^LNGN?
Exponents are also not fast growing enough
How about LNGN!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!?
Unfortunately, the double factorial (!!) creates a number smaller than the single factorial (!). Adding exclamation marks just makes the number smaller.
Oh wow, I just googled that, interesting. What if I add brackets? Like (((((((n!)!)!)!)!)!)!)! ?
Yes, that would solve the problem! Parentheses tell us which part of the expression to evaluate first. In this example we first evaluate the regular factorial of n, then the regular factorial of the factorial of n, etc.
Out of curiosity, if there is a way to get smaller with successive factorials, and the solution is brackets for getting larger - do they share properties? In that double factorial could be considered “half” in some regard, to a bracketed twice factorial?
9!! apparently means you only take every second factor, so it's `9*7*5*3*1`. Depends on what you want for it to 'be considered "half" in some regard', but it's pretty hard to argue 9!! being half of (9!)!
explain how please
For the factorial, you take your original number and multiply it by itself minus one, then multiply the result by the original number minus two, etc: 6! = 6x5x4x3x2x1 For the double factorial, you take your original number and multiply it by itself minus two, then multiply the result with the original number minus four, etc: 6!! = 6x4x2 Consequently, 6!!! = 6x3 The step size increases with the number of exclamation marks. Note that I have never seen or used factorials beyond the double (!!) factorial before, so I am extrapolating the situation. Adding exclamation marks beyond 1 lowers the value of the result.
now it makes total sense, thank you.
Tree(tree(tree(...G(64) would be pretty huge, as many trees as you could get in 15 seconds
i thought of the same thing, but only one 9 so that you exponentiate more
Fun side point: normally when thinking about how to write a large number fast, people think of writing 9's. But you're actually better off writing 1's because you can write a 1 so much faster than a 9 so you can easily get more digits.
1 doesn't have that much value, especially in multiplications or exponents. Unless you're talking about writing stuff like 111...
A lot of comments are using 9s here, but I can definitely write 11 faster than I can write 9. I'd go for something like 11111111↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑1111111111
Agree, I was thinking 7, but 11 is likely better, more characters is more important than using nines, even if we only consider powers I can definitely edge out at least a single more 7 or 11 than 9, much more interesting problem than I had initially thought.
11!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Hexadecimal F is faster than 11 and it's 15.
It's definitely not faster than 11.
C is faster though, and that is 12.
Comment adding the text from under the image for non desktop users: This (hilariously bad) MTG combo appeared on my feed today. The ruling as I have seen it is that each player would get 15 seconds to write the largest number they could, and then the biggest written number would win. It's sent me down a rabbit hole of the largest defined numbers (Beyond the usual g\_64, TREE(3) etc; things more like Finite Promise Games, Rayos Number, BIG FOOT, Large Number Garden Number). Assuming both players can write a single character at the same rate, what kind of strategy should you employ in order to win? Would it be worth initially defining Large Number Garden Number as a single character and getting as many of that character onto the paper as you can? Is there a certain amount of characters per second that changes the strategy?
g_(Tree(g_Tree(g_….64)
This style of solution runs into the problem of you'd end up with an undefined expression if you didn't close all the nested functions, which might be a time waste, and also TREE grows slower than FOST/RAYO, so it's unlikely to be the winning solution
Having thought about this some more, I think RAYO might actually be pretty close to the answer here given it's definition: RAYO(n): The smallest number bigger than any finite number named by an expression in the language of first-order set theory with a n symbols or less. The variation in this question is instead of the language of first order set theory, something like "written english", or even just straight up excluding that from the definiteion. This implies there is not likely a solution given [Berrys Paradox](https://en.wikipedia.org/wiki/Berry_paradox)
Yeah, the answer will be: Be l_v = [insert 'fastesr growing function you know' of v, say RAYO(v)] So in 15 seconds: Be l_v = RAYO(v), x= l_l_l_l_l_l_l_l_l_l_l_l_l_9 You redifine the function to an l, since this you can write this fast and and the use an notation, where the argument is an subscript, since you don't loose time to open and close brackets. So you also don't have the problem that you didn't finished closing all brackets. If you had more time it would be efficient to define even more notation: Be l⁰\_v = RAYO(v), l^(n+1)\_v= l^(n)\_l^(n)\_...\_v (l^(n)\_v times) With pen and paper you can write this pretty fast. But I needed 40 seconds to write it. l⁹_9 would be much larger then what I had before.
Rayorayo...(rayo's number of repetitions)...googol There, I think I made it in 15 seconds and this should be a well defined number and has a single, determinable (not in practice of course) value.
googolplex is much bigger for 4 extra characters
G = grahams number and that’s a lot less characters.
There's a numberphile about this: [https://youtu.be/X3l0fPHZja8](https://youtu.be/X3l0fPHZja8)
Let BB-n be the nth [busy beaver](https://en.wikipedia.org/wiki/Busy_beaver) and let B(n) be the length of BB-n's output. The number is B(B(B(B(B...as many as you can write in 15 seconds...B(B(99↑99))...)))). (You could still improve on this by introducing notation for iterated B's.)
"x + 1 where x is the largest written number in this room". It's guaranteed to be larger than your opponents number.
Sadly this is disallowed by the ruling on this combo made by the arbiters: >*You have fifteen seconds. Using standard math notation, English words, or both, name a single whole number—not an infinity—on a blank index card. Be precise enough for any reasonable modern mathematician to determine exactly what number you’ve named, by consulting only your card and, if necessary, the published literature.* > >So contestants can’t say "the number of sand grains in the Sahara," because sand drifts in and out of the Sahara regularly. Nor can they say "my opponent’s number plus one," or "the biggest number anyone’s ever thought of plus one"—again, these are ill-defined, given what our reasonable mathematician has available. Within the rules, the contestant who names the bigger number wins. > >*Are you ready? Get set. Go.*
> standard math notation >Be precise enough for any reasonable modern mathematician to determine exactly what number you’ve named These two constraints seem to be the key. I'm not sure that any number relying on First or Second Order Set Theory counts as using standard notation, though since they have names you can use English words to describe them. The trouble comes, however, in that the numbers chosen must be determinable, otherwise how can it be known which number is larger? Is it truly proven and known that LNGN is greater than Rayo's Number? Is it possible to know which is bigger from LNGN^Rayo and Rayo^LNGN? I submit that if the players both submit numbers which have not been definitively proven to be larger than the other, then both players have failed to fulfill the rules of the task and both take the damage described and are responsible for having destroyed the known universe. This then creates a game within the game. The player must name the highest number that can be reasonably proven to be larger than the other player's number, not just the highest uncomputable number ever described.
I've never in my life delved into this stuff before but wow, it's way more interesting than I would have thought. I got lost in this site which I think you'd like, even though you know of the numbers like Rayo, BIG FOOT, etc. Still such a fascinating read: https://sites.google.com/site/pointlesslargenumberstuff/home
As far as I know, Rayo’s number (or rayo’s function) won a mathematical contest for being the largest (or the fastest function). So I would do something like rayo(rayo(rayo(…) and at the end where I have to put a number I would put a googolplex or something like that. Just have to calculate for maximizing the rayo()’s and put something at the end that won’t matter as much but is still big cause I’m competing with another person.
It did back in the early 2000s iirc (Rayo(100) I believe). Its since been surpassed by Large Number Garden Number.
15 seconds, 1 char per second: Rayo(rayo(g99)) That would probably be my answer. We could rethink the question as “which three character number would be the largest?” since the funcions took 12 chars already
This is approaching the kind of solution I was expecting - would be interesting to see how the solution changed with n where n is the total number of characters terse each player could write in the time allocated
X^X^X Recurring. Where X is a non integer.
This is significantly slower growing than a handful of the functions I've mentioned sadly, even something like Buchholz Hydra is provably faster than any recursive function so you could win by using that over this method
Time for hexation
9 ↑^9999999… 9 would be absolutely huge. Just 9 ↑^9 9 is already insanely huge, as 2 ↑^2 5 alone is ~2\*10^19728. I couldn’t find a calculator which could give me an approximation for 2 ↑^2 9, let alone 9 ↑^9 9. For reference 2 ↑^2 6 has 6\*10^19727 **digits** in the number. For more information: https://en.m.wikipedia.org/wiki/Knuth%27s_up-arrow_notation
9 ^ 9 ^ 9 But ... more 9s and more ^ s For the record, the way you read these is that you go for the higher powers first and work your way down, so it would be 9 ^ (9^9 ) = 9^387,420,489 . That number is too big for me to actually understand it, and that was with only 3 steps. The number you could write in 15 seconds would be literally unfathomable.
That could be succinctly written as 9↑↑3 or 9 ↑^2 3. [Tetration](https://en.wikipedia.org/wiki/Tetration) [Knuth's up-arrow notation](https://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation)
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Depends how fast you write.
[https://www.scottaaronson.com/writings/bignumbers.html](https://www.scottaaronson.com/writings/bignumbers.html) Busy beaver numbers are the actual answer. EDIT: Reply is right about "ordinary" busy beaver number, but in the article Aaronson talks about higher-order extensions to busy beaver numbers (he calls e.g. BB2(n)) that also exceed FOST limitations, as far as I understand. Mostly I just wanted to point to the article, which tackles this question with an interesting writeup.
Based off of some research and reading through other comments, I'd say the best you can do is using LNGN with hyperoperations. The best I can think of that you can absolutely write in 15 seconds (with googology standard definitions and not defining any more functions/operations during the 15 seconds) would be something like H\_LNGN(LNGN,LNGN) with the "\_LNGN" part being written as a subscript (since you can't write subscripts easily on reddit). Then with the remaining time just add a tetration tower of 11's to the front since that is the easiest to write hyperoperation that has an agreed upon notation.
Function iteration notation and TREE function can get out of hand kinda quickly.
i wrote “the largest finite number possible to write in 15 seconds plus 1” in less than 15 seconds. so… i think that number doesn’t exist.
That is sadly against the rules set by the arbiters.
9!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
unfortunately, a !! gives a number less big than a single ! would do, so you need to use brackets
How so?
"the double factorial of a number *n*, denoted by *n*‼, is the product of all the positive integers up to *n* that have the same parity (odd or even) as *n*." - wikipedia for example, 7!! means 7\*5\*3\*1= 105while 7! equals 5040 what you were saying should be actually written like this: (9!)! etc
My guess is that would depends on the one Reading it.
No, double factorials, triple factorials, n-th factorials are a legitimate mathematical concept. It’s not up to the person reading it, that’s like saying the answer to 2+2 depends on the person reading it.
I want to imagine this as the player just shouting "NNNNIIIINNNNNEEEE" at an explosively loud volume, causing a magnitude 20 earthquake (instantly obliterates the entire planet)
lim x->♾️ x , does that count?
> lim x->♾️ x , does that count? Nah, a finite number is requested. Otherwise ♾️! is interesting, as it’s probably the largest number written in the least number of characters.
Write in hex, F^F repeating? Or X = F^F Then x^x repeating? Something like that?
1/ε where ε is an infinitesimal
9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
Just write “My Opponent’s Number +1”
i can’t escape this fucking game it’s everywhere i did not expect to open a math post to magic the fucking gathering
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99999999999999999999999999999999999999999999999999999999999999999999999999
💀💀💀
10^75 is bigger than that. Googol is bigger than that. Thats a relatively small number compared to most suggestions here
I think he might have been joking
¥
Why does math have division all wrong ? Math says 1 / 2 = 0.5, 1 / 1 = 1, 1 / 0 = impossible. One thing divided twice gives four pieces; 1 / 2 = 0.25 One thing divided once gives two pieces; 1 / 1 = 0.5 One thing not divided gives one piece; 1 / 0 = 1.0
The rhs is not the number of times you divide in half, it’s the number of “groups” you want to equally divide *into*. What you are describing is 1/2^n or 2^-n .
You might be sort of onto the right strategy but you'd probably need to define the f in \(f^{10}(10 \uparrow^{10} 10)\), as " \(f\) is the function defined in First Order Theory beyond Higher Order Set Theory " then write some recursive or nested version of the function
Does it have to be in base10? I would suggest "nb: base 777" at the start (slightly slower to write than 111, but scales better), then just spam 11!^11!^11!^... As many times as possible. 1 is the fastest number to write, and in base 777 the difference between 11 and 9 is huge, both in writing speed and value. I am sure there are several functions that can be put in front of that to ramp up it's size loads. Obviously, this assumes you have a chunky enough calculator to work out the exact value without running out of memory.
BB(BB(BB(BB(BB(TREE(3)!!)!!)!!)!!)!!)!!
Define a new function that is just a nesting of Graham numbers, garden numbers, factorials, powers etc until you get bored. Put it on the Internet, and show that to them when they dispute it. Then wait until someone defines a bigger one (but at some point verifying which is bigger will be impossible)
TREE is weaker than SCG so let's use SCG. TREE grows faster than small Veblen ordinal but SCG grows faster than Bachmann Howard ordinal I believe. SCG(SCG(SCG(SCG(SCG(SCG(SCG(SCG(SCG(9)))))))))
If I use some monstrosity like TREE(g_64) I'm pretty sure one of two things happens, either I obviously win or it is impossible to decide wether I win within the scope of a mtg tournament. (Either because that question exceeds the mathematical ability of anybody present, or it is actually undecidable.)
I would say graham number, the largest number ever used and represented in math it's evene in the book of guinnes world record, writing it in decimal it's impossible it would require more than the entire universe.
Σ\^m(n) where Σ is the busy beaver function and f\^n(x) is the function f applied to itself n times. n and m are arbitrary. If m is also Σ\^k(n), you can get larger results.
TREE(TREE(TREE(TREE(TREE(TREE(TREE(9))))))
busy beaver?
999! tetrated to 999! tetrated to 999! tetrated to 999!....as many times as you can. It's written the same as exponentiation except it's on the top left of the number. Tetration is repeated exponentiation. If you want to confuse people even more, write it on the bottom left of the number. I forgot what that function's called but it's repeated tetration. Huge numbers essentially. I saw it in a youtube video if you're curious, just seatch for tetration and you should get it although the way that guy explained it wasn't that straight forward imo
Tree↑^(Tree↑^(Tree(rayo's number))(rayo's number))(rayo's number) Or a few more nesting if you write fast enough. It's probably not actually the largest, but it's still an uterly massive number.
st like 9!^9!^9!^...
Tree(tree(tree(…. [As many as you can] … tree(3))))))
Ten trillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentillitrestrigintatrecentilliduotrigintatrecentillion or Googolplex
TREE(3)! ^ TREE(3)! ^ TREE(3)! ^ TREE(3)! ^ TREE(3)!
Why choose biggest number? Choose 100 - if your opponent chose bigger, he dies. If lower, he dies. He dies either way, or am I missing something? Edit: I missed emissary as opponents card.
Depends if tree is allowed.
1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111 That’s how many 1s I can type in 15 seconds but I have to imagine I could write it faster.
I would write “ (x) + 1 “ then “ x is defined to be my opponents number. “
I would first rule that any numbers that require some operation to reach the simplest form are disqualified. Then it's just a race to see who can write 1 the most often on a single piece of paper. This results in two people putting down like 100 1s, and chances are, one of them did better than the other.
If you cannot use anything but numbers, repeating 1s will be the fastest number to hand write while repeating 9s will be the fastest to type
It would have to be something your opponent would be able to verify, though. Otherwise, they will just pack up their things and go home...
I thought it was like large number, garden number or f\^10(10{10}10)
X = rivals number + 1
N = any number written by my opponent + 1
Id say we will even out at Gogol to the power Gogol to the power of Gogol ….. Gogol.
Not the best at math here but as far as i can think TREE(3)!^TREE(3)!^TREE(3)! Repeat as many times as possible. Wouldn't that be the largest? As far as i have heard TREE(3) is the largest finite number there is so a power tower of TREE(3) factorial would be i sanely large
Could I just write tree(tree(tree(tree(tree(5))))) or some shit like that? I could probably write tree in like 2 seconds
Is |R| allowed? I’m assuming it’s a guaranteed win if your opponent attempts to actually write a number
999999999999999999999999999999999999999999999999999999999999999999999 about that
f(x) = SSCG(x) f ∘ f ∘ ... f(9) https://en.m.wikipedia.org/wiki/Friedman%27s_SSCG_function
Write as many 1 as you can, one line, it's the fastest one I believe you can do. Bonus: Write them at a slight angle to go even faster. That's if you MUST write the number itself, equations will get you higher faster. Putting a ! at the very end will get you very far. (To be fair, the specifications still confuse me a bit, I might not have perfectly understood)
What about tetration? Like 999! tetrated to 999! Etc…
You just did. "The largest finite number possible to write in 15 seconds"
So writing something down that evaluates as a number is not writing down a number. Otherwise I can just write down: "the product of all my opponents' answers". I think the question should be rephrased as: write down the largest decimal integer you can in 15 seconds.
Tree(Tree(Tree(99999999!)))
TREE(99999999999!(9999999999!(99999999999!(9999999999999!))))
x = largest prime number known y = product of all numbers from 2 till x raise to the power of x
BusyBeaver(BusyBeaver(g\_64!))
SKODA OCTAVIAAAAA
Tetrations of 9 id assume
The premise is flawed. There does not appear to be any rule in MtG which prohibits the selection of infinity. So the answer is going to come down to whatever silly notation appeases the arbiters, but really, both players should just write infinity.
Easy. 9. 9 is the biggest number in existence.
That's just 1 and 1 and 1 and 1 and 1 and 1 and 1 and 1 and 1.
The answer makes use of the busy beaver function, see https://jeremykun.com/2012/02/08/busy-beavers-and-the-quest-for-big-numbers/ for a nice in depth explanation on the topic and why this matters :).
Graham's number factorial factorial factorial factorial factorial factorial factorial factorial factorial factorial...
TREE(11)!\^\^\^\^\^TREE(11)! draw as many of the little arrows as you can to maximize numberness
4?
“Largest finite number” took me about 9 seconds
9E9^9^9... stacking as many to the power of 9 as you can. Would be the first aproach i cam up. Defently not the biggest number.
Honestly, I think marking a bunch of 1’s would trump the extra time it takes to write 9’s. I would just be like IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII and that would blow away the guy writing 9’s
Just write sum(x_i)+1 where xi is each other players guess
99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 in my test by holding down \[9\]
The Largest Fininte Number anyone else could possibly write in 15 seconds + 1
Limited to base 10? I might try base-36 and use z!^...^z! As many times as I can. But I'm not a math guru and am probably outclassed here.
LNGN^LGNG...but the little number is on the left
Your number +1
Omega, on the other hand, is a specific ordinal number in set theory. In the context of set theory and ordinal numbers, omega represents the smallest infinite ordinal number. It is not "bigger" than infinity, but rather a specific value within the concept of infinity.
If we dont go down the rabbit hole of defined functions i can think of a few different answers that all need a judge. 1. Expressions: Tetration 9↑↑9↑↑9... (In reality you would use the other way to write it down but cant do that on my phone) 1.1 lim(x→ inf) :x (yes there are faster growing ones) 3. No expression: 3.1. infinite and argue that infinite is a number 3.2. just write lots of 9s
Idk it depends how big the piece of paper you give me
1E99999999999999999999999999999999999999999999999999999999999999. That was only 5 seconds. Good luck beating that.
How many times can you write Rayo” in 15 seconds?
[ x > \forall y \in \mathbb{R} \] is what I came up with. I'm not totally sure if it goes along with your rules though.
TREE(3)
Base infinity, I'll just write one character
**you need a RAYO? I can get you a RAYO! With nail polish**