On 2014-05-17 19:52, Robert J. Hansen wrote:
Point, but I think it's equivalent: whether it's a flipflop getting a
signal or a microcapacitor that's charging/discharging, in both cases
previous state is getting obliterated and the entropic cost accrues.
:)
Absolutely, no argument there. In fa
> However, the word "normally" is not quite apt. What you normally call
> the RAM of your computer is DRAM, and DRAM is implemented by a charge on
> a capacitor. This achieves much higher densities on a chip than SRAM,
> but is also slower.
Point, but I think it's equivalent: whether it's a flipfl
> The (cold) system where the calculation is done and the (hot) system the
> result is transferred only exchange negligible energy and entropy.
Go build this system, demonstrate you can break Landauer, and collect
your Nobel. Seriously.
___
Gnupg-users
On 2014-05-17 15:28, Robert J. Hansen wrote:
Another way of looking at it: RAM is normally implemented as a
flipflop.
I think the register bank in a processor is still implemented as
flipflops, and all computation ends up there (on a register machine)[1],
so your statement is correct in that
> nt-Type: text/plain; charset=ISO-8859-1
>
> > Now where did you calculate that from?
>
> $dS = \frac{\delta Q}{T}$
>
> Second Law of Thermodynamics, which you just broke. Have a nice day.
>
The (cold) system where the calculation is done and the (hot) system the
result is transferred only ex
> I admit this is beyond my knowledge, but maybe the following is rather
> intuitive and not too incorrect.
Another way of looking at it: RAM is normally implemented as a flipflop.
(The EEs insist on calling them "bi-stable multivibrators," [1] but I
think that's just too kinky for a family-frien
(This mail originally got dropped by the list managing software because
I had accidentally misused a new webmail plugin. I'm resending it
with all original identifiers so it hopefully threads correctly. I'm
also completely ignoring section 3.6.6 of RFC 2822, but who cares? ;)
---
I suddenly
On 17/05/14 01:12, Leo Gaspard wrote:
> Well... If the operation the bit just underwent was a bitflip (and, knowing
> the
> bruteforcing circuit, it's possible to know that), the bit was a '0'.
I admit this is beyond my knowledge, but maybe the following is rather
intuitive and not too incorrect.
