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Old 2017-04-20, 04:57   #46
Romulan Interpreter
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Jun 2011

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Originally Posted by ET_ View Post
Even applying a reduction of a factor 100,000 using hypothetical "optical computers", we need more than 27,6 million years to perform it.
I don't know where you got this from.

Assuming (a) an iterative test method (like Lucas Lehmer, fastest primality test known to man) is found for this type of numbers (10^n+7), and assuming that (b) you have a computer which is a trillion times faster than the actual computers (that is, 10^12, faster than every optical or quantum dreams), and assuming that (c) everybody on the planet (10 billion people in the year 2200, or 10^10) have a billion such computers (yes, each person on the planet, including children and grandmothers, have 10^9 such computers in his/her kitchen), and (d) they exchange data instantly, and assuming that (e) a multiplication algorithm is found which is a million (10^6) times faster than actual FFC. (these all sum to 10^37)

For comparison, such "system" will do a 332M (100 mega-digits) LL test (done by ultra-modern, 10 physical cores (20 hyperthreaded) computers in 60 days) in just a millionth part of a yoctosecond, or micro-yocto-second (more exactly: 1.93*10^-30 seconds). It could do all the job GIMPS did from its inception to present (say 25 years, 100 teraflops continuously), in just about 4.5 zeptoseconds.

And it would test all 50 million primes below 1000000000 in just 9.5 picoseconds.

Do you think this is fast enough?

Note that this is 10^9, all GIMPS' range of interest, without doing any TF/P-1 factors elimination, just do bulk LL for all exponents. The remaining job (with all exponents factored we have today may take GIMPS another 50-100 years or so, including Moore's Law).

Such system will do in one second (a blink of an eye) one hundred billion times all the job that takes humanity 200 years to do.

Say more, that you invent a method to store in few atoms and quants all digits of \(10^{10^{1500}}+7\), and you can instantly access them, and do one iteration of it every yoctosecond (10^-24, or one trillion of trillions of iterations per second).

Then, you still have to do about \(10^{1500}\) iterations, which will take \(10^{1500-24}\) seconds.

Or \(\frac{10^{1500-24}}{3600\cdot 24\cdot 365}\) years.

That is about 10^1468 years....

For comparison, 27 million years is something like 10^7.

Last fiddled with by LaurV on 2017-04-20 at 05:44
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