3 ms·
Your UPDATE is right, attacker can probe a..z 2 times, and just choose the letter that was compressed in both of them, ignoring random compressions
by homakov 13y ago
Your UPDATE is right, attacker can probe a..z 2 times, and just choose the letter that was compressed in both of them, ignoring random compressions
- gojomo 13y agoThanks, but can you clarify... does that mean probing (a..z)×(a..z) (one pair per probe), so there's at least a giant increase in probing required per character? And perhaps even more each character in, since probing for the Nth character now requires (a..z)^(N-1) × (a..z)×(a..z) ? (I'm guessing also, though, it may be possible to probabilistically probe multiple ranges of the secret at once... in a process that seems vaguely similar to forward-error-correction coding.)
- homakov 13y agoalso we need a reflector with 'value="' in the beginning https://twitter.com/homakov/status/364872768921165824/photo/1 https://twitter.com/homakov/status/364872768921165824/photo/...