Zeno's Paradox and Flechette Defense

[smacksim]

goomeister, another voice of math. Yeah!

I think it is interesting that everyone, including me, is trying to clarify and narrow the problem after they have solved it! We know that certainly some part of this is an independent event, but what part?

CT brought up the point that (apparently) ODPs are lost after being committed to stopping an inbound ballistic. Its been a long time since I've had a PB war, so I don't remember, but that does ring a bell. Could someone clarify the mechanics of ODPs and flechettes? Yeah, yeah, I know, it's a simple problem and we should be done with it, but several more interesting variations have come up, so it would be good to recall how PBs are handled. Maybe I'll go try it out and report....

Anyways, the premise that:

If it is impossible for one PB to attack, it is impossible for any number to attack, because each PB attack is an independent event.

can be reversed to state that it is impossible for any single ODP to stop an infinite attack, therefore some attack will get through no matter how many ODPs there are.

The assumption has been that the attack is looked at from the mechanics of a single Planet Buster attempting to poke a hole in a perfect defense. There is no reason why it cannot be framed, or programmed, as one ODP handling x attackers, then moving on to the next ODP.

If you discard this premise, you must come up with some alternate mechanics for the problem. What if we didn't know the answer? What if it were impossible to know before-hand how it would be resolved? If we have n and m quantities we must only accept solutions in which the order of calculations is independent of the result, otherwise we are dictating the mechanics to fit the solution. As CT has earlier shown, quantities below infinity are non-zero, and not surprisingly, approach 1/e for the Planet Busters, while the ODPs have even better odds ...

So we have two kinds of questions:[list=1][*]Scenarios in which we start with infinite quantities.[*]Scenarios where we start with numerical quantities and build towards infinity, and variations.[/list=1]

Surprisingly, there are several opinions on the first type of problem, and noone has written the book on the second type yet either.

[MrWhereItsAt]

Originally posted by Ari Rahikkala
...you can't test it with the methods of physics.

Unfortunately your statement simply means this is all of even less real relevance.

Also, Zeno's Paradox only seems to consider one of the variables involved, distance, as infinitely divisible. I'm not sure whether it compounds matters by dealing with both important parameters (distance and time) as infinitely divisible, or reduces the paradox to, well, not a paradox.

And besides, the uncertainty principle dictates a limit of sorts on the measurement of distance, as it does on time. Thus it is totally meaningless to speak of distances/times smaller than about the value of Planck's constant, which whilst beyond our reach, is in no way infinitesimal.

[Chaos Theory]

Originally posted by goomeister
Hm, interesting problem. I can't resist a good mathematical workout, so let's take a crack at it, shall we? The problem boils down to "What is the probability of m PBs successfully attacking a base with n ODPs, as m and n approach infinity?" So, we need to find the success formula in terms of m and n.

So, let's consider the special case where m=1. Against one ODP, the probability of success are 1/2. Against two ODPs, the probability is 1/4. Against three ODPs, the probability is 1/8. A PB has a 1/2 probability of passing each PB, so the probability that it will succeed against n ODPs is (1/2)^n.

Now, we can derive the formula for m PBs. The chance that a PB will fail against n ODPs is 1-(1/2)^n. Thus, the chance that m PBs will fail is (1-(1/2)^n)^m. The chance that they will succeed is one minus the chance of failure, or 1-(1-(1/2)^n)^m.


P(m, n)=1-(1-(1/2)^n)^m.

With this formula in hand, we can solve the original problem, by finding the limit of P as m and n approach infinity.

lim P(x, x)    =    lim (1-(1-(1/2)^x)^x) = 0
x->+infinity        x->+infinity

Here you made an error. You assumed that m = n. With that assumption, your calculations are correct. However, consider the case where m = 2^n. Let both approach infinity, and then work your calculations.

Therefore, the probability that infinitely many PBs will penetrate infinitely many ODPs is zero.

Think about it this way: the probability that a PB will pass by an ODP is 50/50, a coin toss - heads, the PB survives, tails, it is caught. You have to flip a coin for every ODP, and get heads each time to successfully attack. Now consider infinitely many ODPs: you'd have to flip infinitely many coins and never get tails. But, if you flip a coin infinitely many times, eventually, after a finite number of flips, one will land tails! Thus, one PB can never penetrate infinitely many ODPs. And if the probability of success for one PB is zero, the probability for two PBs is also zero, and for three, and four, and so on up to infinity. If it is impossible for one PB to attack, it is impossible for any number to attack, because each PB attack is an independent event.

But it is not impossible for a PB to get through, its odds simply approach zero. Even though your result is valid, given your assumption, your reasoning as to why this is so is flawed. Considering the case m = 2^n should help with this.

[goomeister]

Grrr... problems this simple shouldn't be so difficult!

I think it is interesting that everyone, including me, is trying to clarify and narrow the problem after they have solved it! We know that certainly some part of this is an independent event, but what part?

CT brought up the point that (apparently) ODPs are lost after being committed to stopping an inbound ballistic. Its been a long time since I've had a PB war, so I don't remember, but that does ring a bell. Could someone clarify the mechanics of ODPs and flechettes? Yeah, yeah, I know, it's a simple problem and we should be done with it, but several more interesting variations have come up, so it would be good to recall how PBs are handled. Maybe I'll go try it out and report....

Very good points. It appeared to me that the attacks were independent. Consider two PBs and one ODP: whether or not the first PB hits, there's still one ODP for the second PB to contend with. This was my reason for assuming the attacks to be independent. However, if ODPs are lost after stopping an attack (and I have a sneaking suspicion that they are), this cannot be so - if one PB is blocked, the next one has a greater chance of success. Is it possible to separate these events?

Here you made an error. You assumed that m = n. With that assumption, your calculations are correct. However, consider the case where m = 2^n. Let both approach infinity, and then work your calculations.

For the purposes of the original problem, it seemed safe to assume that the production of PBs and ODPs was linear. In that case, the limit is zero.

Actually, I thought about this earlier today - if you want some constant nonzero probability P, what must the relationship be between m and n? For the limit to approach a nonzero value, the number of PBs must increase exponentially with respect to the number of ODPs. My back of the envelope calculations indicate that P approaches 1 when m=e^n.

It is impossible to calculate the limit until you know the relationship between m and n. But, if the ODPs are lost after stopping an attack, there is no simple relation between the two - n varies not only with m, but with the number of failed attacks. How can you calculate that, anyway? It is impossible to know the number of failed attacks beforehand, only the probability that an attack will fail. I wonder, is it even possible to solve this problem?

[smacksim]

My adding of probabilities above is completely wrong, but I won't remove it as CT quoted it. Sorry to confuse the issue for the less stat inclined

It must be done like this dice example

[Chaos Theory]

Originally posted by goomeister
It is impossible to calculate the limit until you know the relationship between m and n. But, if the ODPs are lost after stopping an attack, there is no simple relation between the two - n varies not only with m, but with the number of failed attacks. How can you calculate that, anyway? It is impossible to know the number of failed attacks beforehand, only the probability that an attack will fail. I wonder, is it even possible to solve this problem?

I'll assume you're talking about the problem of launching M PBs on a single turn against N ODPs. It's not too difficult, if you know how to solve it:

Originally posted by Chaos Theory

To accurately represent M PBs against N ODPs, first assume that M <= N, otherwise a PB will get through every time. Now, consider X, the number of ODPs that a PB will penetrate before being stopped. X is a random variable with an exponential distribution across the integers, such that P(X) = .5^X (my notation here might be screwy, but I hope it gets the idea across). The expected value (or mean) of X is the sum from 1 to infinity of X*P(X) = x/(2^x). This series has a limit of 2. Therefore, the mean of X is 2. By other calculations, the standard deviation of X is also 2.

The number of ODPs that M PBs will penetrate is represented by M samplings of X. As M -> infinity, this will be a Gaussian distribution with mean 2*M and standard deviation 2*sqrt(M). To find out the odds of whether at least 1 of M PBs will penetrate N ODPs, calculate

(2M - N) / (2*sqrt(M))

and compare the value to a normal distribution chart (or use a function I don't care to find).

[livid imp]

God, are you nerds still arguing math? Why can't you discuss politics or philosophy like normal rambling posters?

[Natalinasmpf]

Sour grapes

[smacksim]

Its so much more satisfying that there might be a solution in mathematics, whereas in politics we can only discover more problems

[Mr. Harley]

Guys, I hate to rain on your boat, but as an old assembler programmer, there is a very simple answer. As your chance PB success is halved, it will eventually reach the point where rounding error exceeds the actual chance. If the smallest number the system can represent is let's say 1%, or .01, you reach that point as you drop below 1%. At this point the processor or the program are going to say that it always equals 1%, or it always zero. If it is zero, the PB never gets through. If it is 1% then eventually it gets through. Obviously it's going to occur at a much smaller quantity, but I am feeling much to lazy to research it and type that many zeroes (to make my point). While mathemiticians could program your calculations into it, trust me for things like Windows and these games, they won't and the rounding issue will prevail. Great discussion though, I've never did get past Calculus 3 (required for Physical Chemistry).

[Natalinasmpf]

trust me for things like Windows and these games,

I use the Linux version, ha!

The C language should have a good grasp of infinite values and the such. Well maybe not, but when we get to quantum computing...

[Ari Rahikkala]

shawnmcc: I'm a young C programmer, and I don't really think the creators of the game bothered to calculate any actual probabilities here. I mean, my computer creates ten million pseudorandom numbers quite happily in half a second (I tested that), so it's not like there's a need to optimise this when no sane person's ever going to build more than a couple of dozen ODPs. Just test if rand() returns a value lower than RAND_MAX>>1 (on little-endian hosts ) as many times as there are defending ODPs.

If we were looking for a really simple answer, we could just declare that playing an infinite amount of turns takes an infinite amount of time...

[Mr. Harley]

Oops, you're right. It just tests each time, so it's always 50%. I still tend to think running statistics, my wife is in a doctoral psych program and I've run some of the stats for her. Unfortunately I've never updated my skills, and most of my programming was in Fortran and Cobol which are obsolete for very good reasons. So it's just a simple statistical regression, in which case you're testing for the PB not getting through - the desired result. Which has nothing to do with the infinite case, which I will leave to those very capable people on this thread.

[Natalinasmpf]

This brings up infinity:

What exactly IS infinity?

Would it be possible for computers what humans perceive to be infinite as finite?

Is it simply because of the mathematical limitation of our minds?

[goomeister]

Originally posted by Natalinasmpf
This brings up infinity:

What exactly IS infinity?

Would it be possible for computers what humans perceive to be infinite as finite?

Is it simply because of the mathematical limitation of our minds?

Infinity is a quantity such that, for any real number n, infinity>n. Also, a set S is infinite if there exists a proper subset of S which has the same cardinality (size) as S.

Infinity is not a real number, so operations that work with real numbers do not work with infinity. This is why we use limits: 1/infinity is not a real number, but we can say that 1/n approaches zero as n approaches infinity.

I'm not sure what you mean by computers "percieving" infinity - computers do not percieve anything. If you mean, "Does infinity exist in reality, or is it just a mathematical tool?", that's a matter of philosophy. Personally, I don't think it matters - if the methods work, there's no reason to doubt the underlying concepts.