Crypto and new computing strategies
Jim choate
ravage at bga.com
Wed Mar 30 12:48:36 PST 1994
>
> >I am not shure that it has been demonstrated that a QM mechanis is necessarily
> >solely of a Turing architecture.
>
> The Bekenstein Bound gives limits both on the expected maximum number
> of quantum states encodable in a given volume of space and on the
> expected maximum number os transitions between these states. If this
> bound holds (and it certainly seems to hold for EM fields), then a
> probabilistic Turing machine will be able to simulate it.
>
> >Also there is the potential to use neural networks at these levels (which are
> >not necessarily reducable to Turing models, the premise has never been proven)
>
> If you have infinite precision, the statement is unproven. If you
> have finite precision, you get a Turing machine. You never get
> infinite precision in real life, even with quantum superposition.
>
> Steve Smale did some work a few years ago where he made Turing-type
> machines out of real numbers, i.e. infinite precision. P=NP for this
> model, and the proof is fairly easy. From an information-theoretic
> point of view, you can encode two real numbers inside of another one
> and do computations in that encoded form, because a real number
> encodes an infinite amount of information.
>
> If it's finite, it's a Turing machine. If it's expected finite, it's
> a probabilistic Turing machine. If it's infinite, it cannot be
> implemented in hardware.
>
> Eric
>
First off, EM fields are NOT QM. They do have some characteristics which
'bleed' over form the Quark level. Also since EM fields are made of
hardons and not leptons (which an electron is) may blow a hole in this
approach since leptons do not follow the same sort of charge conservation
rules as hadrons.
As to infinite precision and its non-presence....Beeep....wrong answer...
Electrons change state in zero time, this implies at least some form o f
infinite precision (otherwise how does the system know the difference between
zero and some small-o value?). I suspect this is another error based on
the implied (and incorrect) implication in this line of discussion that
hadrons and leptons use the same rules.
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