On 04/09/2010 15:07, victor.ducho...@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] --------------------------------------------------------------------- The Cryptography Mailing List Unsubscribe by sending "unsubscribe cryptography" to majord...@metzdowd.com