[ExI] for the fermi paradox fans
anders at aleph.se
Mon Jun 16 20:56:26 UTC 2014
Robin D Hanson <rhanson at gmu.edu> , 16/6/2014 11:18 AM:
I finally had a chance to look over these papers, and I'm not convinced they are modeling the interesting case. With reversible computers there is of course a lot of thermal noise, but the system is designed to robustly accommodate noise in the flexible state (eg local charge) of the system, though not of course in the structure (eg conductor vs insulator) that channels those flexible states. In real computers the main noise other than thermal noise in the flexible state is due to cosmic rays. But that noise is very highly correlated spatially, which makes it much easier to deal with; just redundantly do big computation chunks in spatially separated places and vote on the answer.
I think the flexible states are more easily disrupted than the structure states (since they are flexible): whatever the cause of errors are - thermal motion, cosmic rays, occasional tunneling - it ought to mess up the flexible parts of the system far more often than the structure. Hmm, this might go for reservoirs of negentropy too: a bit error in a negentropy reservoir is a slight depletion. It would be nice to have a proper theory of the expected resiliency of systems like this so we could make quantitative predictions.
Redundant computing with votes still has to be reversible. A voting gate selecting the majority of A, B and C will be dissipative unless it outputs the majority vote plus two bits of information that allows reconstructing ABC. And while the non-faulty chunks can then be rewound and re-used, there has to be error correction for all the bits in the noise-affected one. So while the error rate has gone down from p to p^2, if a single module error occurs there will still be the same number of bits to clean as in the single-module case. So the overall amount of bits that need to be cleaned per unit of time goes from Np to 3p((1-p)^2 + p(1-p) + p^2)N=3p(1-p+p^2)N - the far lower error rate comes at a price of extra error correction. (if I calculated things right).
I wonder if there is a way to make something like a computer gate that can switch to energy collector mode when a cosmic ray comes through. Then the noise might perhaps more than pay for itself.
But how do you know that the ray will be arriving?
I have a suspicion that one can make a Maxwell's daemon argument that noise is useless to power computation. However, real radiation might be different since it is not at maximum entropy.
Anders Sandberg, Future of Humanity Institute Philosophy Faculty of Oxford University
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