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From: Stathis Papaioannou <stathispapaioannou.domain.name.hidden>

Date: Thu, 21 Apr 2005 23:01:46 +1000

Jesse,

I've deleted everything, it was getting too messy. I hope this

(semi-)mathematical formulation captures your argument correctly:

Suppose you start with one individual, your friend, on a computer network

which has infinite resources and will grow exponentially forever. This

individual will be duplicated every unit time period, so that after n

generations, there will be 2^n copies. Suppose there is a probability p(n)

that, during a unit time period, any single individual in the nth generation

will be effectively deleted, whether through suicide, murder, local hardware

failure, or whatever. To simplify the maths, assume that if at least one

copy survives, the rest of the 2^n copies in that generation will be

restored from memory; but if all the copies are deleted, then that will be

Real Death for your friend. Now, the probability of Real Death in the nth

generation, given the above, is p(n)^(2^n). If p(n) is a constant, call it

k, this probability will clearly decrease with each generation. The

probability that your friend will never suffer Real Death is then given by

the infinite product:

(1-p(0)^(2^0))*(1-p(1)^(2^1))*(1-p(2)^(2^2))*...

which I believe converges to a value between 0 and 1 (too lazy to work it

out now) and is another way of making your point, with the geometric series,

that even with an always-nonzero probability that a given individual will

die, if this probability is always decreasing due to exponential growth of

copies of the individual, the probability that at least some copies will

survive indefinitely does not limit to zero.

Now, look at p(n) again. This time, let's say it is not k, but a random real

number greater than zero, smaller than 1, with k being the mean of the

distribution. At first glance, it may appear that not much has changed,

since the probabilities will "on average" be the same, over a long time

period. However, this is not correct. In the above product, p(n) can go

arbitrarily close to 1 for an arbitrarily long run of n, thus reducing the

product value arbitrarily close to zero up to that point, which cannot

subsequently be "made up" by a compensating fall of p(n) close to zero,

since the factor 1-p(n)^(2^n) can never be greater than 1. (Sorry I haven't

put this very elegantly.)

The conclusion is therefore that if p(n) is allowed to vary randomly, Real

Death becomes a certainty over time, even with continuous exponential growth

forever. If you have a real world network, or simulated sentient beings, I

don't believe it is possible eliminate the random lement in this parameter.

--Stathis Papaioannou

_________________________________________________________________

Update your mobile with a hot polyphonic ringtone:

http://fun.mobiledownloads.com.au/191191/index.wl?page=191191polyphonicringtone

Received on Thu Apr 21 2005 - 09:20:34 PDT

Date: Thu, 21 Apr 2005 23:01:46 +1000

Jesse,

I've deleted everything, it was getting too messy. I hope this

(semi-)mathematical formulation captures your argument correctly:

Suppose you start with one individual, your friend, on a computer network

which has infinite resources and will grow exponentially forever. This

individual will be duplicated every unit time period, so that after n

generations, there will be 2^n copies. Suppose there is a probability p(n)

that, during a unit time period, any single individual in the nth generation

will be effectively deleted, whether through suicide, murder, local hardware

failure, or whatever. To simplify the maths, assume that if at least one

copy survives, the rest of the 2^n copies in that generation will be

restored from memory; but if all the copies are deleted, then that will be

Real Death for your friend. Now, the probability of Real Death in the nth

generation, given the above, is p(n)^(2^n). If p(n) is a constant, call it

k, this probability will clearly decrease with each generation. The

probability that your friend will never suffer Real Death is then given by

the infinite product:

(1-p(0)^(2^0))*(1-p(1)^(2^1))*(1-p(2)^(2^2))*...

which I believe converges to a value between 0 and 1 (too lazy to work it

out now) and is another way of making your point, with the geometric series,

that even with an always-nonzero probability that a given individual will

die, if this probability is always decreasing due to exponential growth of

copies of the individual, the probability that at least some copies will

survive indefinitely does not limit to zero.

Now, look at p(n) again. This time, let's say it is not k, but a random real

number greater than zero, smaller than 1, with k being the mean of the

distribution. At first glance, it may appear that not much has changed,

since the probabilities will "on average" be the same, over a long time

period. However, this is not correct. In the above product, p(n) can go

arbitrarily close to 1 for an arbitrarily long run of n, thus reducing the

product value arbitrarily close to zero up to that point, which cannot

subsequently be "made up" by a compensating fall of p(n) close to zero,

since the factor 1-p(n)^(2^n) can never be greater than 1. (Sorry I haven't

put this very elegantly.)

The conclusion is therefore that if p(n) is allowed to vary randomly, Real

Death becomes a certainty over time, even with continuous exponential growth

forever. If you have a real world network, or simulated sentient beings, I

don't believe it is possible eliminate the random lement in this parameter.

--Stathis Papaioannou

_________________________________________________________________

Update your mobile with a hot polyphonic ringtone:

http://fun.mobiledownloads.com.au/191191/index.wl?page=191191polyphonicringtone

Received on Thu Apr 21 2005 - 09:20:34 PDT

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