# Tag Info

The problem with the HMAC-based solution you drew up is if the shared secret $s$ has low entropy; for example, it's actually a password that could conceivably be in a dictionary. In this case, someone could listen to the exchange $r_1, \operatorname{HMAC}(C \mathbin\| r_1, s)$, and go through his dictionary of possible values of $s$, and see if any one of ...