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There are several open source GNFS tools that can do 1024 very efficiently on GPUs, and even cheap consumer GPUs have 10s of thousands of cores now, even by you
by DarkmSparks 2y ago
There are several open source GNFS tools that can do 1024 very efficiently on GPUs, and even cheap consumer GPUs have 10s of thousands of cores now, even by your measure "2700 core-years" is only around a month or so on a single consumer grade GPU.
Not "free", but any malicious actor has access to a lot more than a single GPU.
The UK government also has several huge arm based solutions dedicated to cracking internet encryption, zero chance that isn't breaking mostly everything, for sure the Chinese and Russians have similar.
- SahAssar 2y ago> The UK government also has several huge arm based solutions dedicated to cracking internet encryption, zero chance that isn't breaking mostly everything, for sure the Chinese and Russians have similar. So you seriously think that almost all current RSA is being decrypted in real time by at least UK, China and Russia (and I would assume US)? Do you have any source or reference for this at all?
- upofadown 2y ago>There are several open source GNFS tools that can do 1024 very efficiently on GPUs, ... Reference? Why has no one demonstrated this?
- littlestymaar 2y agoYet nobody is collecting the RSA-numbers bounties? RSA-270 (much, much easier than 1024 compute-wise) has a bounty of $75k, why would it be unclaimed then when you can spend three years worth of cloud rented H100 (I'm being conservative here and count $3/h which is far from the best deal you can get) and still make a profit? Also a GPU core and CPU cores really aren't comparable individually, so your “consumer graphic card having thousands of core already” is comparing apples to oranges.
- DarkmSparks 2y agohad a bounty. the RSA challenge unexpectly finished in 2007, task is left to the reader to speculate what happened in 2007.
- littlestymaar 2y agoOh, I wasn't aware of the end of the challenge. But 1024 was definitely not broken by then, at least not by brute force.
- DarkmSparks 2y agonone of it is "brute force", GNFS is a process that rapidly excludes numbers from the search space that cannot be the answer, in principle similar to the way they broke enigma. numberphile has a great video on that one https://www.youtube.com/watch?v=V4V2bpZlqx8 https://www.youtube.com/watch?v=V4V2bpZlqx8 Also, taking the OP as a "worse case", afaik: 512bit = $8 so 1024 = 8^2 = $64 2048 = 8^2^2 = $4,096 4096 = 8^2^2 = $16,777,216 noting $8 for 512 seems very expensive to me.
- deleted 2y ago[deleted]
- btdmaster 2y agoThat's not correct. Consider, for example, a processor that can handle 2^31 computations per second. 2^32 operations can be computed in 2 time units, whereas 2^64 operations will take 2^33 time units. search_space(n: number_of_bits) = 2^n * k so search_space(1024)/search_space(512)=2^512, not 2^2. Asymptotics in GNFS are better[0], but only on the order of e^(cbrt(512 * 64/9)) times more work, not 2^2. This would give an approximation of math.exp(math.cbrt(512 * 64/9))*$8 = $40 million for 1024 bits. [0] https://en.wikipedia.org/wiki/General_number_field_sieve https://en.wikipedia.org/wiki/General_number_field_sieve
- DarkmSparks 2y agoPretty sure the search cost of GNFS is (bits)^2, the search cost of brute force is 2^(bits), if it was 2^(bits) GNFS would be no better than brute force. ->but only on the order of e^(cbrt(512 * 64/9)) e^(log(n)) = n