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Sep
14
awarded  Scholar
Sep
14
accepted Is there a theoretical maximum useful keysize given the block-size?
Sep
11
comment Is there a theoretical maximum useful keysize given the block-size?
Hi @lxgr -- I'm with you as far as understanding that there are $2^M!$ distinct permutations, and therefore $2^M!$ keys as well. But when it comes to representing those keys, seems to me that $\lceil \log_2 2^M! \rceil$ bits should be sufficient -- permutations can be lexicographically ordered, and the $n^{th}$ permutation can be deduced quickly using the Factorial Number System. Of course, this is still an insanely large number.
Sep
11
comment Is there a theoretical maximum useful keysize given the block-size?
Hi @CodesInChaos -- Liked the $\log_2 2^N!$ answer. I need an integral number of bits, so it will be $\lceil \log_2 2^N! \rceil$ if I can tolerate some key collisions and $\lfloor \log_2 2^N! \rfloor$ if I can't. Why don't you put your comment as an answer so that I can accept it? I'm a newbie here, so can't up-vote comments...
Sep
10
awarded  Student
Sep
10
asked Is there a theoretical maximum useful keysize given the block-size?