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Jul
31
comment How do I extract the private key components $N$ and $D$ from a private RSA key?
Also, this doesn't directly relate to your question, but the above key is a 512 bit RSA key. Keys of that size have been broken, it is generally recognized that everyone needs at least 1024 bit keys (with 2048 bits being highly recommended). I'm pointing this out in case this is the key size you were intending to use. OpenSSL will be glad to generate larger keys, if you ask it to.
Jul
31
comment How do I extract the private key components $N$ and $D$ from a private RSA key?
Actually, it's not all that specific to OpenSSL; OpenSSL uses the standard DER-encoded format (as mandated by PKCS#1), base-64 encoded (because DER-encoding uses all 256 byte values).
Jul
28
comment OTT service using FPE
@e-sushi: actually, to answer the question "is it possible", the important thing to ask is "does the network that carries the Codec-encoded bitstream just move those bits from the sender to the receiver, or does it do any manipulation (such as doing some internal compression to save bandwidth on an internal link). If the system is just moving bits, I see no problem in sending encrypted bits -- if the system expects that is an actual signal (and so transforms that don't change the signal audibly are allowed), that makes the problem much trickier.
Jul
28
reviewed Leave Closed Understanding Lattice based cryptosystem
Jul
28
comment Why not the one-time pad with pseudo-number generator
Am I missing something? In step 4, you append the seeds to the ciphertext, and apparently send them in the clear. Why can't an attacker use those seeds to decrypt the message? If the answer is "because the attacker doesn't know the 'non-standard' constants in the key streams", well, that means that those non-standard constants are effectively the key to this system; using keys bolted into the system makes for, at least, very questionable security.
Jul
25
reviewed Close Key Derivation from random salted seed, HMAC and HMAC based KDF
Jul
25
reviewed Close Are there any Javascript CSPRNGs?
Jul
25
reviewed Leave Open Elliptic curve group over a prime finite field $F_p$
Jul
24
comment Vigenère cipher: Security when key length and plaintext length are the same
@tylo: I can't think of anything else anyone could add (except for possibly the mention of Venona as an example) -- why don't you convert your comment into an answer?
Jul
24
reviewed Close Secure way to store fixed size string of digits
Jul
23
reviewed Leave Open Why is MAC using nonce+message+hash(nonce+message+identifier) not the standard?
Jul
23
comment Different padding rules for Merkle-Damgard and Keccak/sponge function
@CGFoX: you can't have two distinct messages that pad into the same. For example, if one message (in binary) was 101101 and the other was 10110100, the first would pad to 101101100000, and the second would pad to 101101001000 -- they don't look the same.
Jul
23
comment Different padding rules for Merkle-Damgard and Keccak/sponge function
@CGFoX: it is (and, in fact, they do -- they just add a length field as well). However, I believe that the original MD designers may have been afraid that wasn't sufficient -- that if somewhere were to find a fixed point, then it would allow a collision (and the length field would invalidate that).
Jul
23
comment Different padding rules for Merkle-Damgard and Keccak/sponge function
@CGFoX: we always add precisely one 1, and then add zeros; if the input ended with a '1' bit (or '0x01' byte), we'll still pad by adding a '1' bit (or, if we're dealing with bytes, an 0x80 0x00 0x00 ... 0x00 pattern)
Jul
23
answered Different padding rules for Merkle-Damgard and Keccak/sponge function
Jul
23
answered Textbook RSA with exponent e=3
Jul
23
reviewed No Action Needed Which algorithm do you recommend for practical use to generate unique passwords for each website?
Jul
23
comment How is this affine function a pair wise independent permutation?
@sashank: no; the probability that $f(A)=C$ is $1/2^n$; $f(B)$ is independent of $f(A)$ except we know that $f(B)\ne f(A)$, hence given the first holds, there are $2^n-1$ possible values that $f(B)$ can take on. It is equidistributed other than that, and hence joint probability that $f(A)=C$ and $f(B)=D$ is $1/2^n \times 1/(2^n-1) = (2^{2n}-2^n)^{-1}$
Jul
22
comment RSA with modulus n=p²q
Like ECIES; decryption is a lot faster than RSA; for encryption, you can do precomputation so you can do most of the work before you see the message.
Jul
22
answered RSA with modulus n=p²q