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5

I have argued so 15 years ago, and not been proven wrong since. Basically, A5/1, with a $n$-bit state, offers a resistance of roughly $2n/3$ bits of security. With $n = 64$, the resistance is very low, thus amenable to not only direct breaking, but also all kinds of trade-offs. All the attacks published so far are dances around that resistance level of ...


3

No, this mode as listed does not provide integrity of the decrypted plaintext. An active attacker can flip arbitrary bits from the first 64 bits of the decrypted plaintext freely without causing a decryption failure. He can do this by modifying the $iv$; the decrypted $tag_0$ will authenticate (because the $iv$ is not used to compute that), and then the ...



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