Sujin, Your statement “after attacker has cracked a CGA address, then it is in the same stand as the defender” is incorrect. The attacker will not need 65536 additional trials; he will need to find a matching CGA address 65536 times, and each trial is just as complex as finding a matching CGA address. The defender on the other hand can pick any CGA address he wants, so he has to perform just 65536 trials.
-- Christian Huitema From: zhou.suj...@zte.com.cn [mailto:zhou.suj...@zte.com.cn] Sent: Wednesday, March 28, 2012 4:46 AM To: Christian Huitema Cc: ipv6@ietf.org; Jari Arkko Subject: 答复: RE: about security level evaluation of draft-zhou-6man-mhash-cga-00 Regards~~~ -Sujing Zhou Christian Huitema <huit...@microsoft.com<mailto:huit...@microsoft.com>> 写于 2012-03-27 23:00:09: > > Well, I think it is quite a simple trade-off. Increasing Sec > increases computational effort on both sides by equal amount. > Increasing the length of > > the hash increases computational effort only on the attacker side. > As a result, the hash bits are relatively valuable. > > Jari, the effect of sec is different on both sides. The effort is > not increased "by equal amount" but rather "in the same proportion." > Suppose that an attacker could crack the 59 bit hash in a day with > sec=0. Adding sec = 1 means that the attacker will have to try 65536 > times as many keys. The time to crack becomes 65536 hours, i.e. > about 7 years. The defender will have to try 65536 instead of one to > get the right sec, taking only a fraction of a second. after attacker has cracked a CGA address, then it is in the same stand as the defender. defender, as well as attacker, has to try one by one to get the right sec, i.e., to get a matching length of zeros, how come attacker need extra 65536 hours but defender only needs a fraction of a second? > > 1) Status quo: RFC 4982 eats three bits for Sec, leaving 59 bits > of hash length after accommodating for u and g bits as well. The > good with this is > > that it leaves maximum size for hash. The bad is that it is less > flexible for adding many new hash algorithms. > > I kind of like the status quo. > > > 2) The proposed new approach: Eat a total of six bits, for Sec and > the hash algorithm identifier. 56 bits remain. The good is that this > gives more > > flexibility to allocate hash algorithms, including the ability to > independently choose Sec and the algorithm. The bad is that the hash size is > > decreased. > > Decreased a lot. On the other hand, if we do gain the flexibility to > define new algorithms, then we can define a mandatory-to-implement > algorithm that incorporates the equivalent of the "sec" strengthening. the security of a hash algo can be roughly estimated by length of its output, from birthday attack, that is about 2^(59/2) or 2^(56/2)(theorical estimate), or less, can you give a more precise of value of "a lot"? I don't see flexibility in defining mandatory-to-implementing hash alg > -- Christian Huitema > > > >
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