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Jan
9
suggested approved edit on openssl tag wiki excerpt
Jan
9
comment How to check the signature without the predetermined function in ECDSA from openssl?
This question (and your followup) might be better suited for Stack Overflow. Crypto.SE is really more for the theoretical side of cryptography (i.e. we'd be happy to explain how ECDSA works, but "how to use OpenSSL" is getting a bit off topic). You may want to see our help center for more information about our scope.
Jan
9
comment Why IV does not have to be secret yet has to be random
This is a nice and well-written answer, but alas, it doesn't seem to answer the actual question asked here, which is (as far as I can tell, the question being somewhat unclear) specifically about CBC encryption and its requirement that IVs be unpredictable. Still, have a +1 anyway. :)
Jan
9
reviewed Reject How much stronger is RC4 if it is keyed with an RSA private key?
Jan
9
comment Do we have anyway of knowing if avoiding self-permutation in RC4 makes it any stronger?
... However, the RC4 internal state also includes the $i$ and $j$ indices, which are initialized to fixed values; thus, the full RC4 state space has $256^2 \cdot 256!$ states, of which only $256!$ (those with $i=j=0$) are possible on the first round (and some, like the Finney states, are provably unreachable on any round).
Jan
9
comment Do we have anyway of knowing if avoiding self-permutation in RC4 makes it any stronger?
@IamNick: All the 256! permutations can, indeed, be reached; to show this, first note that a Fisher-Yates shuffle can produce any permutation, and then that any execution of the Fisher-Yates shuffle on {0, ..., 255} can be simulated by RC4 key setup with a suitably chosen key. In fact, this directly yields an algorithm for reconstructing the key based on the final permutation, which is exactly what I've presented above. [...]
Jan
9
comment Why can’t the public key exponent in RSA be negative?
Good point, @Meysam. I've edited my answer to correct that.
Jan
9
revised Why can’t the public key exponent in RSA be negative?
added 465 characters in body
Jan
9
answered Why can’t the public key exponent in RSA be negative?
Jan
9
comment Why can’t the public key exponent in RSA be negative?
While not precisely a duplicate, at least my answer to this related question also answers yours.
Jan
9
revised Do we have anyway of knowing if avoiding self-permutation in RC4 makes it any stronger?
link to Finney's original sci.crypt post
Jan
9
revised Do we have anyway of knowing if avoiding self-permutation in RC4 makes it any stronger?
added 752 characters in body
Jan
9
answered Do we have anyway of knowing if avoiding self-permutation in RC4 makes it any stronger?
Jan
9
comment Do we have anyway of knowing if avoiding self-permutation in RC4 makes it any stronger?
Technically, since RC4 key length is capped at 256 bytes (and the key is repeated if it's shorter than that), just concatenating $k_{ip}$ and $k_u$ won't work; you'd have to actually run the key setup twice, first with $k_{ip}$ and then with $k_u$. Still, for any initial permutation $p_1$ and any target permutation $p_2$, there is an RC4 key (possibly several, in fact, but at least one is easy to find) that transforms $p_1$ into $p_2$. In that sense, every key in the OP's modified RC4 is provably equivalent to some key in normal RC4.
Jan
8
revised Permutation of keys that guarantees different hashes
deleted 15 characters in body
Jan
7
comment Rule 30 Cellular Automaton for Cryptography
This paper seems to say "no, it's not secure", but I haven't read it fully yet.
Jan
7
comment Why IV does not have to be secret yet has to be random
You may also find this question useful.
Jan
7
answered Permutation of keys that guarantees different hashes
Jan
7
comment Permutation of keys that guarantees different hashes
@fgrieu: I suspect the OP actually wants $$\exists H,m\ \forall x,y: (x \ne y \land H(x) = H(y)) \implies H(m(x)) \ne H(m(y)),$$ rather than $$\exists H,x,y\ \forall m: x \ne y \land H(x) = H(y) \land (m \ne {\rm id} \implies H(m(x)) \ne H(m(y))).$$
Jan
7
comment Permutation of keys that guarantees different hashes
@petermlm: Yes, but if $H(x) = H(y)$ for all $x$ and $y$ (that's what $\forall$ means), then the output of $H$ is constant.