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Sha256 vulnerability for full rounds

github.com

16 points by giosch 9 years ago · 11 comments

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jey 9 years ago

Same thing that's explained by https://crypto.stackexchange.com/a/48586 ?

TLDR: It's easy to find fixed points of hashes like SHA-256.

  • simias 9 years ago

    Ah, great link. I was not aware of this "feature" of SHA256:

    >To abuse this property you need to get the state of the hash to match a state you get when running the decryption of the blockcipher underlying the compression function. Finding such a match requires a meet-in-the-middle attack with cost 2n/2 and thus isn't cheaper than finding a collision.

  • amenghra 9 years ago

    laie makes it sound like they found two things (free-start collision attack and circular hash attack).

    I agree the free-start part isn't very interesting but I don't think we have enough information to confirm or dismiss whether the circular hash attack part is novel.

    • dsacco 9 years ago

      Using that philosophy, we have to be cautious about whether or not the Riemann hypothesis has been solved every time someone uploads a paper claiming a proof to arXiv, even if it's nonsensical (which this "proof" is), just because we haven't had a team of mathematicians peer review it.

      Let me assure you: there is nothing novel here. There is no vulnerability. You want to start from a place of skepticism with these things, not a place of, "Well we don't have enough information to say it's not true..."

    • simias 9 years ago

      Extraordinary claims require extraordinary evidence, I think it's fair to dismiss those claims until they're capable of coming with a better justification than "I developed an entirely new type of cryptanalysis theory to achieve this."

    • jey 9 years ago

      Could you unpack "circular hash attack"? Googling was not very helpful.

      • amenghra 9 years ago

        "circular hash attack" left me confused and waiting to hear the full story. I totally agree with "Extraordinary claims require extraordinary evidence".

grovegames 9 years ago

How much of a concern is this? Do we now need to use SHA512 for everything, or is this more of an academic vulnerability that we won't see in the wild?

amenghra 9 years ago

It's unfortunate that proof.py doesn't give an example of a message block that leads from h0 to the q._h constant.

I.e. Free-start collisions don't let you create two PDFs with the same sha256 hash.

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