"Infinite" life

[Maveric78f]

Once more, that's a theoretical question. It has nothing related to real tourney games, even if it could happen in a very strange MU.

Well, we all know that infinite combo do not exist in magic, one has to say I repeat this sequence of actions N times. So let's say my opponent is playing life combo. He says he activates nomad 10^100 times on Daru and sacs it to Diamond Valley. He gains 2*10^100 life and he's now at 20+2*10^100 life.

Aftewards I play the strange combo: Opalescence + Doubling Seasons + Followed Footsteps. With Opalescence and Double Seasons in play I can enchant Doubling Seasons with Followed Footsteps, so that at my next upkeep I'll get 2 more Double seasons (3 DS). Then, the next upkeep, 8 more (11 DS). Then, the next upkeep 2^11=2048 more (2059 DS). At my fourth upkeep, I'll get 2^2059 + 2059 DS. And I'll attack my opponent for the win, the turn after.

The number of DS after the nth upkeep which we'll note DS(n) follows this simple rule:
DS(0)=1
DS(n+1)=DS(n)+2^DS(n)
As we've seen, it become very big, very fast. So fast, that for n>10, you don't have enough paper in the world to write down the number.

Now, let's say that my opponent understood that and he comboes at game 2 but he can't think how to name the number of times he repeats the targetting on Daru, so that he'll still be out of range of my non-infinite combo. Does it mean that he can't win more than what my combo will eventually generate? Or can he use mathematical functions to express this number of (and basically say "I target DS(10^100) times Daru") ?

[Dark_Shakuras]

Have to use an english number. Lucky for you, there is Nongentillion. That's 10^2703.

[Maveric78f]

Have to use an english number. Lucky for you, there is Nongentillion. That's 10^2703.

Definitely not enough. Actually, you can't write it down without a sequence.

[Rico Suave]

First to answer your question, your opponent can name a Googolplex, or even a Googol. These are english numbers much in the same way a dozen refers to 12.

A googolplex will suffice to gain more life than anything you can possibly take away. To give you an idea of just how big it is, if we were to write a googolplex out on paper there would not be enough space in the known universe to hold all the paper.

Well, we all know that infinite combo do not exist in magic, one has to say I repeat this sequence of actions N times.

This is wrong. There are two kinds of "infinite" combos, or loops, in magic:
1) Loops with at least one optional action.
2) Loops with only mandatory actions.

The only time you say "repeat this X number of times" is when there is an optional action that can break the loop (such as the player can choose not to target the Daru with Nomad).

If the loop is comprised only of mandatory actions the game results in a draw. For example if I have a Worldgorger Dragon in my graveyard and cast Animate Dead on it, there will be a loop that follows which will continue on and on with only mandatory actions. The only way to break this loop would be to eventually target another creature in the graveyard with the Animate Dead, but if there are no other creatures in the graveyard aside from Worldgorger Dragon then the loop goes on indefinitely and the game results in a draw.

[Maveric78f]

I know that Rico Suave. But I just think that you can't imagine how big this sequence DS(n) is.

googolplex=10^(10^100)=10^googol

I recall that DS(4) = 2^2059 + 2059.
As 2^10 > 10^3, DS(4) > 10^200 = googol²
%%%EDIT%%% Actually this last line could be DS(4) > 10^600 = googol^6 %%%
DS(5) > 2^(googol²)
DS(5) > (2^googol)^googol

So that DS(5) >>> googolplex

[tivadar]

I think you broke my head Maveric... But it's a valid question. There's also the question of how one would go about counting/keeping tokens on the offensive side. That's a lot of tokens. And not to mention, with it growing that fast, as we've pointed out, you wouldn't have enough paper on which to write out how many tokens you have.

I think my answer is "yes, you've broken magic, now please stop" :-P. You're right, this situation should/would never actually come up. I'm guessing there's no actual rule for it, nor does there need to be.

But still, good question!

EDIT: If someone proves me wrong, awesome. Obviously MTG is a game for rules junkies :-P.

[Maveric78f]

I think you broke my head Maveric... But it's a valid question. There's also the question of how one would go about counting/keeping tokens on the offensive side. That's a lot of tokens. And not to mention, with it growing that fast, as we've pointed out, you wouldn't have enough paper on which to write out how many tokens you have.

I think my answer is "yes, you've broken magic, now please stop" :-P. You're right, this situation should/would never actually come up. I'm guessing there's no actual rule for it, nor does there need to be.

But still, good question!

Yeah maintaining the game state is impossible with this combo.

[cdr]

When you gain an arbitrary amount of life, you do have to end up with an actual number. You can represent it however you like, however.

I would be fine with someone using exponential notation or even a function to represent an aribtrary life total, especially when it's being used to stay out of range of a non-infinite combo.

As far as tracking a gamestate that silly goes, you'd have to write the number of tokens down, again likely as a function.

[tivadar]

Yeah maintaining the game state is impossible with this combo.

Though I will say that in theory, the Life.dec player should win. They're working in the larger infinite...

[Maveric78f]

When you gain an arbitrary amount of life, you do have to end up with an actual number. You can represent it however you like, however.

I would be fine with someone using exponential notation or even a function to represent an aribtrary life total, especially when it's being used to stay out of range of a non-infinite combo.

As far as tracking a gamestate that silly goes, you'd have to write the number of tokens down, again likely as a function.

The problem is that my opponent might not understand maths, so that we can't agree on a game state. That's a fundamental problem according to me.

[Forbiddian]

Aftewards I play the strange combo: Opalescence + Doubling Seasons + Followed Footsteps. With Opalescence and Double Seasons in play I can't enchant Doubling Seasons with Followed Footsteps, so that at my next upkeep I'll get 2 more Double seasons (3 DS). Then, the next upkeep, 8 more (11 DS). Then, the next upkeep 2^11=2048 more (2059 DS). At my fourth upkeep, I'll get 2^2059 + 2059 DS. And I'll attack my opponent for the win, the turn after.

The number of DS after the nth upkeep which we'll note DS(n) follows this simple rule:
DS(0)=1
DS(n+1)=DS(n)+2^DS(n)
As we've seen, it become very big, very fast. So fast, that for n>10, you don't have enough paper in the world to write down the number.

Checking the math....

Ok, DS + FF on DS = 2 tokens.

First upkeep: 2 doubling season + 1 token = 2^2 + 2 = 6.

Second upkeep: 6 doubling season + 1 token = 2^6 + 6 = 38


I think it grows faster than you presume, and I can't find an equation to represent it.

DS(n)=2^(n^n + (n-1)^(n-1) + (n-2)^(n-2)... (n-n)^(n-n) + 1) + (n^n + (n-1)^(n-1) + (n-2)^(n-2)... (n-n)^(n-n) + 1)
DS(0)=2
DS(1) = 2^2 + 2
DS(2) = 2^6 + 6
DS(3) = 2^27+6

Crud, that equation doesn't grow fast enough. I could express it as something less convenient, but your equation doesn't hold since you forgot to double the token entering play with doubling season. You start with two doubling seasons out your first upkeep, so you start the next turn with 6.




But at any rate, even if your equation for the number is right: A googolplex is 10^(10^(100)), not (10^100)^2. It's impossibly larger than the puny number you claimed. (10^100)^2 is just 10^200.

A googolplex is ACTUALLY:

10^(10,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000)

Compare 2*10^2 (what you claimed the exponent is) to 10^100 (the exponent of what it actually is).

Yes, your number beats 10^(10^2), but it doesn't beat 10^(10^(10^2))



But that's not even CLOSE to Graham's number, which is 3 to the fourth power tower of 3 and the number I always pick if people ask.

I don't really want to get into the mathematics. I'm sure wiki or something has an article, but the number is so impossibly large.

It makes googolplex^googolplex^googolplex^googolplex^googolplex look like 0.

Even with a quickly diverging equation, like: n^n^n, you won't catch Graham's number unless n is in the googolplexes^googolplexes^googolplexes.

And I don't think you can take even a single googolplex turns during the time limit.



In short: If the person picked a googolplex, you lose. If he picked a googol, you win easily. If he picked Graham's constant, he gave you a quick lecture in mathematics and you walk away an enlightened man.

[cdr]

The problem is that my opponent might not understand maths, so that we can't agree on a game state. That's a fundamental problem according to me.

The life player has to pick a number and an accurate representation. If he can't, that's his problem. If he has the ability, you are out of luck.

[Maveric78f]

In short: If the person picked a googolplex, you lose. If he picked a googol, you win easily. If he picked Graham's constant, he gave you a quick lecture in mathematics and you walk away an enlightened man.

In short, you succeeded to take something mathematically right and add errors in it to prove false things.

Let's start with the first upkeep:
1 double season in play, 1 token becomes 2 tokens. 3 DS in play after the first upkeep.
Second upkeep, 1 oken becomes 2^3=8 tokens. 11 DS after the second upkeep.

I don't even understand what you're trying to do.

My maths are good. Try to figure out how I did it before trying to demonstrate something else.

[Nihil Credo]

I hope Opalescence is in Masters' Edition 4.

[MEATROCKET]

But that's not even CLOSE to Graham's number, which is 3 to the fourth power tower of 3 and the number I always pick if people ask.

I came to this thread to post this, but also to note that Graham's number is probably the best (read: my favorite) number to chose in this case since it makes both players happy. Graham's number (G) is the largest individual number with a name without doing silly things like G squared, or G up arrow G (aw, that's just not fair). So choosing G should satisfy both players in that it's really freaking (incomprehensibly, absolutely mind-bogglingly) large, and that it is also not arbitrary so it can be referenced later without any problems.

[Rico Suave]

I know that Rico Suave. But I just think that you can't imagine how big this sequence DS(n) is.

googolplex=10^(10^100)=10^googol

I recall that DS(4) = 2^2059 + 2059.
As 2^10 > 10^3, DS(4) > 10^200 = googol²
%%%EDIT%%% Actually this last line could be DS(4) > 10^600 = googol^6 %%%
DS(5) > 2^(googol²)
DS(5) > (2^googol)^googol

So that DS(5) >>> googolplex

You started out wrong.

A googolplex is 10^10,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000

10 to the power of a number with a hundred zeroes.

If you have any basic idea of math you already know you would never get remotely close.

[Maveric78f]

googolplex=10^(10^100)

What's wrong with all of you?

You don't have to write down the zeros to make it big. If you don't understand that 2^2059 > 10^100 (with is googol, not googolplex) by far, I can't do anything for you.

[Rico Suave]

What's wrong with all of you?

You don't have to write down the zeros to make it big. If you don't understand that 2^2059 > 10^100 (with is googol, not googolplex) by far, I can't do anything for you.

What's wrong is you started making incorrect remarks like this:

So that DS(5) >>> googolplex

[Maveric78f]

Because that's what I've demonstrated, since googleplex = 10^googol.

OMG. I don't even know I'm talking to. Stop saying you don't agree with the conclusion, just tell which line you can't understand, I'll explain it.

[Forbiddian]

In short, you succeeded to take something mathematically right and add errors in it to prove false things.

Let's start with the first upkeep:
1 double season in play, 1 token becomes 2 tokens. 3 DS in play after the first upkeep.
Second upkeep, 1 oken becomes 2^3=8 tokens. 11 DS after the second upkeep.

I don't even understand what you're trying to do.

My maths are good. Try to figure out how I did it before trying to demonstrate something else.

My bad, I thought that Followed Footsteps also put a token into play immediately (for whatever reason).

It's not altogether important, because my point about your equation being bogus is still valid.

DS(n+1)=DS(n)+2^DS(n)

doesn't represent anything.

If DS(n+1) appears to be the function for the number of DSs in play given the number of DSs in the previous turn. Duh, except that's totally useless as a full equation.

DS(n)=DS(n) is just as useful because we're still far from a closed form representation of the number of doubling seasons in play for a given n. And even more importantly, we're looking for a closed-form equation for the amount of damage dealt to a player, which adds one more layer.

I don't want to do that, but probably if I ran that doubling season combo, I would.




You keep saying "This number is bigger than a Googol, therefore it's bigger than a Googolplex," which is WRONG!

We tried to explain it to you, but this:

I recall that DS(4) = 2^2059 + 2059.
As 2^10 > 10^3, DS(4) > 10^200 = googol²
%%%EDIT%%% Actually this last line could be DS(4) > 10^600 = googol^6 %%%
DS(5) > 2^(googol²)
DS(5) > (2^googol)^googol

Is retarded as shit. Nobody cares about a googol^2. Who would even say that number? Googolplex is massively larger. MASSIVELY larger.

2^2059 is small compared to what we're talking. It's like, 8^500, so maybe 10^400th. A Googolplex is 10^(10^(10^2)). Just looking at the exponent:

400<<10^(10^2)

therefore, 2^2059 << A googolplex.