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I found a hashcat benchmark results in the internet: hashcat results:
Why does SHA-512 take less time? SHA-512 is longer and I thought it therefore needs more time and computation resources to compute the hash?
The only differences in calculations are the initial value and the output size. From NIST FIPS 180-4
- The initial hash value, $H^{(0)}$, shall be set as specified in Sec. 5.3.4; and
- The 384-bit message digest is obtained by truncating the final hash value $H ^{(N)}$, to its left-most 384 bits: $$H_0^{(N)} \mathbin\|H_1^{(N)} \mathbin\|H_2^{(N)} \mathbin\|H_3^{(N)} \mathbin\|H_4^{(N)} \mathbin\| H_5^{(N)}$$
When two source codes are compared inc_hash_sha384.cl and inc_hash_sha512.cl one can see that actually, SHA-384 does less job than SHA-512. It is not coded simply truncate the SHA-512 with different initial values. For the little speed difference, one explanation can be the vectorization that helps SHA-512 during the final calculations.
In the case of password protection, the tiny difference does not make much difference. In password hashing, actually, we use specially designed password mechanisms like PBKDF2 and Argon2 which was the winner of the password hashing competition.
In the security perspective, SHA384 has resistances to length extension attack but SHA512 doesn't have. SHA384 has 128-bit resistance against the length extension attacks since the attacker needs to guess the 128-bit to perform the attack. This is due to the truncation.
The different initial value provides domain separation. With domain separation $$\operatorname{SHA384}(m) \neq \operatorname{SHA512}(m)|_{384}$$ where $|_{384}$ is the truncation.
In the case of pre-image, secondary pre-image, and collision resistance in generic attacks we have;
\begin{array}{|c|c|c|c|}\hline & \text{pre-image resistance} & \text{2. pre-image resistance} & \text{collision resistance} \\\hline \operatorname{SHA-384} & \mathcal{O}(2^{384}) & \mathcal{O}(2^{384}) & \mathcal{O}(2^{192}) \\\hline \operatorname{SHA-512} & \mathcal{O}(2^{512}) & \mathcal{O}(2^{512}) & \mathcal{O}(2^{256}) \\\hline \end{array}
The collision resistance is $\mathcal{O}(\sqrt{2^{n}}) = \mathcal{O}(2^{n/2}) $ due to the generic birthday attacks.
Therefore; SHA-512 is better in the case of pre-image, secondary pre-image, and collision resistance. SHA-512's only weakness is that it is prone to length extension attacks.
Why does SHA-512 take less time?
Not for cryptographically relevant reasons. Only examination of the code (source, perhaps object) could tell. Essentially, SHA-384 is SHA-512 with a different starting constant, then removing 128 bits of the result. One possibility is that this removal is in separate code using the code for SHA-512 as a subprogram. It could also be a code alignment issue, a branch taken vs not taken... Whatever the reason, it is off-topic on crypto.SE.
SHA-512 is longer and I thought it therefore needs more time and computation resources to compute the hash
Yes, SHA-512 is longer; but no, it does not need sizable more time, because internally it operates on a 512-bit result that is truncated in the end. There is almost nothing to save in not computing the 128 extra bits: about 1/4 of the last round out of 80 (that's 0.32%), and two 64-bit additions (make that 0.4%).
