[145811] in cryptography@c2.net mail archive
Re: RSA question
daemon@ATHENA.MIT.EDU (Francois Grieu)
Sun Sep 5 18:09:17 2010
Date: Sun, 05 Sep 2010 12:12:15 +0200
From: Francois Grieu <fgrieu@gmail.com>
To: cryptography@metzdowd.com, cryptography@randombit.net,
Victor.Duchovni@morganstanley.com
In-Reply-To: <20100904130737.GR7575@np305c2n2.ms.com>
On 04/09/2010 15:07, Victor.Duchovni@morganstanley.com apparently wrote :
> one could look-up the brute-force cost of RSA
> vs. (ideal) symmetric algorithms, and discover that
> while brute forcing an ideal symmetric algorithm doubles
> in cost for every additional key bit, GNFS factoring cost
> is approximately proportional to
>
> exp(n^(1/3)*log(n)^(2/3))
>
> where "n" is the number of key bits.
>
No. As far is known, the effort is approximately proportional to
exp( (n*(64/9+o(1)))^(1/3) * log(n)^(2/3) )
Both the 64/9 and the o(1) term change the effort estimate way more than
a constant factor.
> So compared to 1k RSA bits, 16k RSA bits has a GNFS cost
> that is (16*1.96)^(1/3) ~ 3.15 times higher.
This is wrong well beyond the omission of the 64/9+o(1) term. By
straight application of the above formula with n=16384 and n=1024, and
expressing the ratio as the nearest power of 2, we get that the cost of
factoring 16 kbits RSA numbers would be approximately 2^219 times that
of factoring a 1 kbits RSA number if we neglect the o(1) term [and still
2^114 rather than 3.15 times neglecting the 64/9+o(1) term].
> If 1k RSA bits is comparable to 80 symmetric bits, then 16k RSA bits
is comparable to 80*3.15 or 252 bits.
One can't multiply key bit size by ratio of effort; we must instead
*add* the *logarithm* in base 2 of the ratio of effort.
We get that 16k bits RSA would be comparable to 80+219 = 299 bits
symmetric key if we neglect the o(1) term [80+114 = 194 bits neglecting
the 64/9+o(1) term].
Francois Grieu
[Wondering if the crisis of financial institutions may have to do with
how they do math]
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