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Even in a perfect world (the random oracle model) there's no way to ensure first-preimage and second-preimage resistane of more than $2^n$ and collision resistance of more than $2^{n/2}$. (Wie $n$ bytes as the output size of the hash function.) That's the maximum you will ever be able to archive. Cutting off some bytes of the output of a secure hash ...


4

Collision and preimage resistance does not imply this; suppose we select a collision and preimage resistant function $H$ with a known value $I$ with $H(I) = 1$ (the group identity); this additional assertion does not contradict the assumptions of collision or preimage resistance. Then, given $a$, we can easily output the tuple $(I, a)$; as we have $H(I) ...


0

Response to Question 2 : algorithm 1 : change the algorithm into this, it will be easier to explain hash1 = sha512(password + salt1); hash2 = sha512(password + salt2); for (i = 0; i < 1000; i++) { hash1 = sha512(hash1); // <-- Do NOT do this! hash2 = sha512(hash2); // <-- Do NOT do this! } For 2 different inputs hash1 and hash2, if ...


3

This partial answer establishes (rather trivial) lower and upper bounds for the asymptotic hardness of the problem, assuming $h$ behaves like an $n$-bit wide random function. If one hashes $m$ messages $M$, then computes $f(i,j,k)=h(M_i)\oplus h(M_j)\oplus h(M_k)$ for $(i,j,k)\in\mathbb {Z_m}^3$, that's $m^3$ results, with most values duplicated at least 6 ...



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