On Wed, May 01, 2002 at 01:37:09AM +0200, Anonymous wrote: > This is probably not the right way to approach the problem. Bernstein's > relation-finding proposal to directly use ECM on each value, while > asymptotically superior to conventional sieving, is unlikely to be > cost-effective for 1024 bit keys. Better to extrapolate from the recent > sieving results.
Yes, good point. > For about $200 you can buy a 1000 MIPS CPU, and the memory needed for > sieving is probably another couple of hundred dollars. So call it $500 > to get a computer that can sieve 1000 MIPS years in a year. You need a lot more than a couple of hundred dollars for the memory, because you'll need 125 GB per machine. See Robert Silverman's post at http://groups.google.com/groups?hl=en&selm=8626nu%24e5g%241%40nnrp1.deja.com&prev=/groups%3Fq%3D1024%2Bsieve%2Bmemory%26start%3D20%26hl%3Den%26scoring%3Dd%26selm%3D8626nu%2524e5g%25241%2540nnrp1.deja.com%26rnum%3D21 According to pricewatch.com, 128MB costs $14, so each of your sieving machines would cost about $14000 instead of $500. > If we are willing to take one year to generate the relations then > ($500 / 1000) x 8 x 10^10 is $40 billion dollars, used to buy > approximately 80 million cpu+memory combinations. ($14000 / 1000) x 8 x 10^10 is $1.1 trillion. So my earlier estimate for a $10 trillion 4-month machine was only off by a factor of 3, which is a nice coincidence. :) (Too bad about the memory, otherwise you can get your sieving machine almost for free by writing a worm to take over half of the Internet, which currently has about 190 million hosts.) --------------------------------------------------------------------- The Cryptography Mailing List Unsubscribe by sending "unsubscribe cryptography" to [EMAIL PROTECTED]
