This is the last I will be saying on the subject. I am not interested
in teaching a course on thermodynamics.
> Well... A nuclear reactor produces 1GW, and thus produces 1PJ in
> 10^6 s, that is approx. 11 days 14 hrs. Sure, you may be very
> interested in Health & Safety compliance of nuclear re
First: I agree with everything skipped in the quotes.
On Wed, May 14, 2014 at 07:31:26PM -0400, Robert J. Hansen wrote:
> On 5/14/2014 6:11 PM, Leo Gaspard wrote:
> > BTW: AFAICT, a nuclear warhead (depending on the warhead, ofc.) does
> > not release so much energy, it just releases it in a dead
Hello, i'm new to this list/community so I hope this is the place to report
such things.
* Links for list pages are broken in https://lists.gnupg.org/: there's a port
(8002) in the urls which if you remove will take you to the correct pages.
These links are ok in https://www.gnupg.org/document
Quantum cryptography was only discussed relating either to asymmetric
crypto, which AES isn't, or in relation to Grover's algorithm, which is
used to brute-force an algo.
Peter is correct, but a little clarification may be in order.
Grover's is not a brute-forcing algorithm: it's a search alg
On 16/05/14 14:37, Michael Anders wrote:
> In fact arriving at a realistic estimate for the energy needed to brute
> force AES is really hard work. (Besides: Who can say for sure that we
> cannot get some bits from cryptoanalytic progress(two bits already
> crumbled).
You cannot get bits of crypta
Hi all,
I answer my self, after, many many tests done, in fact it isn't actually
possible to do it under sid debian => root cause bug on systemd :
Debian Bug report logs - #618862
systemd: ignores keyscript in crypttab
link here : https://bugs.debian.org/cgi-bin/bugreport.cgi?bug=618862
Best R
> Now where did you calculate that from?
Forgot one more reference -- look at Schneier's _Applied Cryptography_,
where he talks about the physical limits of the cosmos. He has a
physicist's error in his presentation (he's off by a factor of ln 2),
but he confirms the Second Law necessity of a hea
> Now where did you calculate that from?
$dS = \frac{\delta Q}{T}$
Second Law of Thermodynamics, which you just broke. Have a nice day.
And no, I am not going to explain this further. My reason for this is
simple: you need to take college-level courses in differential and
integral calculus, pa
On Wed, 2014-05-14 at 22:26 +0200, gnupg-users-requ...@gnupg.org wrote:
> If you want to run the temperature lower than the ambient
> temperature
> of the cosmos (3.2K), you have to add energy to run the heat pump --
> and the amount of energy required to run that heat pump will bring
> your
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA512
Hi
On Thursday 15 May 2014 at 5:55:08 PM, in
,
Peter Lebbing wrote:
> Decryption using a wrench rather than a key;
> http://xkcd.com/538/ (don't forget the on-hover text!)
I guess I never hovered over the picture before.
- --
Best regards
MFP
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