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It's trivial to calculate collisions because the message blocks M1..Mn can be interchanged without issue. So EK(M1 XOR M2) = EK(M1' XOR M2') where M1' = M2 and M2' = M1. Different messages should obviously not calculate to the same MAC authentication tag.


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This depends on the MAC because there are different kinds of attacks to consider. If the best attack is randomly trying authentication tags, then the key does not matter. If the best attack is brute forcing the key, then key renewal does mean that the attacker has to "start anew", but as long as the key space is large enough that the probability of finding ...


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It seems to me that the answer to the question, as literally asked, is $0$. This is because you're asking for the minimum advantage "for every adversary", and the set of possible adversaries includes e.g. the trivial adversary that simply always outputs an empty message and a random tag $t \in T$, and thus has a probability of exactly $1/|T|$ of successful ...


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I am answering on the basis of this paper (pdf) linked in the comments, as well as some of the related papers it cites or is cited by. I am not aware of more realistic attacks on HMAC. It assumes a DPA side channel that leaks the number of bits flipped when a new value is read into a CPU register (or in another instruction in some of the papers). I.e. it ...


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The problem is with XOR-ing the message blocks together to make a single compressed block. This reduces the message in a highly regular way, meaning it would be very easy to construct collisions. A length extension attack on this MAC design can consist of any data at all provided one of the blocks was the binary complement of the remainder - trivial to ...



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