Re: [tor-talk] NSA supercomputer

[Andrew F <andrewfriedman101@xxxxxxxxx>]

Anthony, good point.  And worth a lot more then $0.02


Thanks Seth excellent write up.  I will have to brake out the sci
calculator and run some number.
I know the flops issue is a big one, but thats the only measure I could
find for the big system in Utah.
However, your point is well taken.  No way to really know without testing.
How about a road trip... we could knock on the the door and ask for 10
minutes of computer time?
Knock knock... "hello Mr NSA, can we use your super secret spy computer for
10 minutes?"
And Yes, My next post after asking that question will be from sunny
Guantánamo Bay.  As I am sure I will get an all expense paid trip
 from our friends in the (*Redacted *).

You know, if anyone has an Nvidia Xk20 and an AMD 16 core working together,
we could test on a small scale and then extrapolate from there, get an
estimate of efficiency per second and do the calculations.  If anyone wants
to mess around with it and has the hardware...  :-)  I'll buy the pizza and
beer. In fact, It would be a fun article to write.    "So just how fast is
the NSA supercomputer?"

Ok, everyone, have a good weekend.








On Fri, Apr 5, 2013 at 9:33 PM, Anthony Papillion <anthony@xxxxxxxxxxxx>wrote:

> On 04/05/2013 01:01 PM, Andrew F wrote:
> >
> > Basically he said that with quantum computing all bets are off and every
> > cipher today will likely be cracked. Quantum computing will require new
> > kinds of ciphers and only those with Qcomputers will be able to decrypt
> the
> > messages.
>
> Not entirely correct, as I understand it. Granted, quantum computing
> will shred most (all?) of the ciphers we currently use. But that's
> mostly because they will be able to do massively efficient prime
> factorization using something like Shor's algorithm
> (https://en.wikipedia.org/wiki/Shor%27s_algorithm). If I understand
> correctly, resisting such technology doesn't require creating a cipher
> that takes a quantum computer to decrypt but one that is resistant to
> efficient factorization.
>
>
> Just my $0.02,
> Anthony
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