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What if text would replaced with H(text)? Will it weaken the HMAC algorithm? Yes. It makes collision attacks on the hash function apply to the MAC, which isn't normally the case with HMAC. You can find a pair $(m, m')$ that hash to the same value, get the MAC for one of them and move it to the other. That means that your modified HMAC construction ...


6

Will it weaken the HMAC algorithm? No, assuming that $H$ is a collision resistant hash function (which is a stronger constraint than what HMAC puts on the hash function). The security property that a MAC (such as HMAC) has is that "given a large set of $message, tag=HMAC(K, message)$ pairs (where $K$ is an unknown key), where the attacker can choose ...


2

I would use HMAC-SHA256. While poncho's answer that both are secure is reasonable, there are several reasons I would prefer to use SHA-256 as the hash: Attacks only get better. SHA-1 collision resistance is already broken, so it's not impossible that other attacks will also be possible in the future. It allows you to depend on just one hash function, ...


1

Any PRNG with a finite state size is eventually periodic. The maximum period possible is $2^n$ for an $n$-bit state, but the average with a well mixed state is $2^{n/2}$. Here the hash function used is SHA-512, but the state is 1024 bits. A first guess would be a period of $2^{512}$, rather than the $2^{256}$ mephisto gives. Let's look at the cycles. Both ...


1

Yes, this PRG is theoretically periodic. Approximately after generating $2^{512}$ outputs a state will be generated that collides with a previous state. (A previous version of this answer said $2^{256}$ as I missed that two outputs are used for the state. Otus answer pointed out this mistake.) This follows from the birthday problem. However, $2^{512}$ is ...


1

This won't seriously impact the security of the key. HMAC is pretty resilient and changing the last part of the hash won't allow attacks on the a hash such as SHA-256. Note that you only select 4 characters, of which the last one only encodes 4 bits (as it is at the end). That means you've got a check value the size of 2^22 encoded bits, i.e. a chance of 1 ...


1

How is the MAC or HMAC secret shared with the other party for verification? It depends on the type of implementation or system you are interested in, but i can share my experience. Unless pre-agreed key is known before hand, usually a public key cryptography is used to share secret key (e.g. AES-key) over a public insecure channel. Encrypting by the ...



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