Comment by wolf550e
7 years ago
What? No, brute forcing 2^256 is not an option. We assume brute forcing 2^128 is not an option. If brute forcing 2^256 were an option, why would anyone use AES-256?
7 years ago
What? No, brute forcing 2^256 is not an option. We assume brute forcing 2^128 is not an option. If brute forcing 2^256 were an option, why would anyone use AES-256?
No. Its not an option in practice. But its the starting point for a research problem.
* 100x different AES-256 keys generated by /dev/random would have 25,600 bits of entropy. (All 100x keys are independent, and would require an attack effort per key).
* 100x different AES-256 keys generated by /dev/urandom would "only" have 256-bits of entropy. (/dev/urandom would be attacked, assuming /dev/urandom started with 256-bits of entropy in its pool, assuming /dev/urandom never got extra entropy in the 100x calls)
You can't pretend that the two solutions to the problem are the same. At best, you can suggest that 256-bits of entropy is enough for practical purposes.
No, brute forcing 2^256 is not the starting point of anything.
Again, after seeding with initial entropy, /dev/random and /dev/urandom are the same, /dev/random is not better entropy.
Attacking a single 256 bit key is exactly as equally impossible as attacking a 256 bit RNG pool used to generate 100 keys.
There is no actual scenario in which using hardware RNG to generate 100 keys would keep me safe while using a hardware RNG to initialize a deterministic RNG to generate 100 keys would lead to compromise.
If your attacker model includes brute forcing 2^256, I can't help you. If you're trying to prove that Aleph-one is greater than Aleph-0, perhaps, but the number of atoms in the universe is less than Aleph-0, so it's irrelevant to cryptography or to information security.
People who say "maybe an advance in math will break 2048 bit DHE" are at least theoretically correct. But no advance in math is going to break that RNG. If you can't trust that RNG, you can't send secrets using CTR, GCM, CHACHA, etc. because they will all break using that method.
> Attacking a single 256 bit key is exactly as equally impossible as attacking a 256 bit RNG pool used to generate 100 keys.
But both are strictly easier than attacking 100x independently created 256-bit keys. The /dev/random case.
> If you're trying to prove that Aleph-one is greater than Aleph-0
I mean, yeah, that's basically the argument I have. 256-bits of entropy are probably enough for your application, but you cannot under any circumstances tell me that its the same as 25,600 bits of entropy. That's what /dev/random vs /dev/urandom comes down to. For most people in most situations, its almost inconceivable to imagine an application that needs more than 256-bits of true high quality entropy.
EDIT: A POTENTIAL case, which I admit I haven't though too much about... is a 2048-bit RSA key. I assume the RSA key needs to be generated with 2048-bits of true entropy (please correct me if I'm wrong), but the bits-of-security are far less than that, because RSA doesn't scale like other encryption schemes. So your 2048-bits of entropy RSA-key only scales to 128-bits of security.
If your 2048-bits of entropy RSA-key ended up to only have 256-bits of entropy generating it, you're possibly in bad luck (only 16-bits of entropy assuming a similar scaling factor). I'm not a specialist in the math, but... not all cryptographic primitives scale 1-to-1 with the input entropy source.
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What are you talking about? This is not an "in practice" detail.
Let's put the time involved in perspective.
``` int main() { for (unsigned long long i = 0; i < (((unsigned long long)1) << 35ull); i++) { __asm__ __volatile__ (""); } } ```
this is large enough to remove the relevance of any start up time, and make noise somewhat meaningless.
On my laptop this takes 9 seconds. That's just counting to 2^35. Now let's imagine we're just counting to 2^256. Not even doing the work to break the rng. How long does it take?
2^221 * 9seconds.
Which is approximately 1.397 * 10 ^ 50 times the age of the universe (I think a billion billion billion billion billion times the age of the universe).
That's just counting from 0 to 2^256, not doing anything else at all.
Understand that if you can break a CSPRNG with 2^256 bits of state, you can break:
* The entire WebPKI system, because all you're doing is brute forcing an very high speed 256 bit hash function.
* TLS connections: you're just brute forcing 256bit keys here as well, in the worst case. Fortunately most are still just aes128. That's 6471949087173059816 times faster, so that's only a billion billion billion times the age of the universe.
Note that the numbers involved mean that "making computers faster" and "adding more computers" doesn't making a meaningful difference, because you run into an energy wall - literally more power than a star emits ends up being necessary.
So, please don't just regurgitate nonsense: brute forcing a current 256 bit CSPRNG is not possible, at all. because physics.
So to break it, you have to break the csprng algorithm. But if you can do that, then you can break any of the places the an RNG is used in practice. Eg. you could have a "true" rng seeding crypto, but it doesn't matter because the attacker can just use your algorithm to break the result of your crypto.
Note also the "true" random you get on a computer is fairly restricted - it has both little variance, and also significant bias, so the output has to be merged (hash function that you've broken) and whitened (again, hash function that you've broken). So even if you were trying to rely on "true" random, you're not getting as much "entropy" as you seem to think you are, and you've invented an algorithm that breaks all the massaging used to make it seem more random.