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0 pointsgruez2y ago0 comments

Generating 2^64 hashes isn't guaranteed to produce a collision, and even if a collision did exist in that set, you're not going to find it by getting a bunch of GPUs to compute 2^64 hashes. There's a huge difference between a haystack that maybe contains a needle, and a needle that's been pulled from the haystack and presented to you. To actually find and identify the collisions you'll need to hook those GPUs up to some sort of storage/retrieval system. Just to store 2^64 128-bit hashes would take 295.1 exabytes. That's an order of magnitude more storage than NSA's utah datacenter[1].

[1] https://en.wikipedia.org/wiki/Utah_Data_Center

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moyix2y ago

You can generate pairs of hashes for random inputs and check for collision without storing all of the outputs, no?

adrian_b2y ago

True, which is why 256-bit or bigger hashes are recommended to make sure that the time for finding a collision is alone great enough to make this impossible.

Finding a collision for an 160-bit hash by brute force, on a big cloud or supercomputer, is about the maximum that someone with unlimited financial resources could do.

gruezOP2y ago

That requires way more hashes to be computed though.

moyix2y ago

I don't think that's correct? It's probabilistic, yes, but in expectation you would still need ~2^64 hashes to find a collision for a 128-bit hash (birthday paradox).

gruezOP2y ago

See my above comment. There's a huge difference between having a collision somewhere in a pile of 2^64 hashes, and actually having the colliding hashes ready to present.

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moyix2y ago

You can generate pairs of hashes for random inputs and check for collision without storing all of the outputs, no?

adrian_b2y ago

True, which is why 256-bit or bigger hashes are recommended to make sure that the time for finding a collision is alone great enough to make this impossible.

Finding a collision for an 160-bit hash by brute force, on a big cloud or supercomputer, is about the maximum that someone with unlimited financial resources could do.

gruezOP2y ago

That requires way more hashes to be computed though.

moyix2y ago

I don't think that's correct? It's probabilistic, yes, but in expectation you would still need ~2^64 hashes to find a collision for a 128-bit hash (birthday paradox).

gruezOP2y ago

See my above comment. There's a huge difference between having a collision somewhere in a pile of 2^64 hashes, and actually having the colliding hashes ready to present.

1 more reply

j / k navigate · click thread line to collapse