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As far as we know, SHA512 acts like a random function. So, the only way we know to find a preimage whose hash starts with 0x12345678, is to go through distinct preimages, and hash each one until we find one that starts with 0x12345678. If the output of SHA512 is equidistributed (and we have no reason to believe it isn't), then the probability of any hash ...


3

If we use $H_1(X) = H_0(X) \oplus firstnbits(X)$, this would seem to be trivial. EDIT: As C├ędric Van Rompay pointed out, this is only a counterexample if $H_1$ winds up being preimage-resistant. This may not be a necessary consequence of $H_0$ being preimage-resistant, but I really only need one case where it is.


2

When the output of the hash function is $n$ bits, then there are $2^n$ possible outputs. For a preimage attack you are given a hash $h$ and you need to find a message $m$ where $h = H(m)$. Since there are $2^n$ possible outputs, the probability of guessing an input that that maps to the given output is $\dfrac{1}{2^n}$. So on average you need to try $2^n$ ...


1

Consider this hash: $$H(m) = m$$ Where we define it's domain to be messages of some arbitrary fixed-length. It is completely second pre-image resistant. It is not at all first pre-image resistant. Therefore: Second pre-image resistance does not imply first pre-image resistance.


1

Assuming the hash is strong, you will not be able to find a preimage any more easily when you know multiple hashes (whether they use the same function or another). In fact, with some hashes you will be able to derive $H(S||K)$ from just $H(S)$ and $K$ (known as the length extension attack), so it can't give you more information. Unless the hash is broken, ...



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