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Jul
25
comment Transforming Gaussian random bytes to uniform
What are the parameters of the gaussian generator? What is the actual distribution of output bytes?
Jul
23
comment Publish my Encryption algorithm and test it
(that said it's commendable that you discovered probabilistic encryption on your own, if you did)
Jul
23
comment Publish my Encryption algorithm and test it
What do you hope to gain by publishing it? Do you seriously believe your algorithm is in any way superior to existing, proven algorithms designed by expert cryptographers? (case in point: pretty much the only feature that a non-cryptographer could capitalize on is speed, and your algorithm is apparently slow)
Jul
21
comment Adapt existing GOST code in C to hash a file
I'm voting to close this question as off-topic because it is about interfacing with a cryptographic implementation and so belongs on StackOverflow.
Jul
10
comment One cipher to rule them all?
Threefish, obviously
Jul
9
comment Where can I find names for cryptographic notation I don't recognize?
Furthermore, if the greek letter has not been defined in the surrounding context then it's likely it's not cryptographic notation at all but some quasi-universal notation that the reader is expected to be familiar with (e.g. $\sum$ for sums, $\prod$ indicating a product). You learn these by doing math, and @djsutho's table can help in that case too (also, if you know it's greek, you can en.wikipedia.org/wiki/Greek_alphabet to find the letter's name and google it)
Jul
7
comment Is one-time-pad still secure if the number of 1's in the key is revealed to the attacker?
Hint: what if you are given that $n = 1$? $n = 2$? How many keys does that rule out, and how many are left? (are they equiprobable?)
Jul
7
comment Diffe-Helman Exchange result is always 1
@Jake Basically, 47 is not a generator modulo 23, but 5 is. It's not called "generator" for nothing, it must have special properties that make it, well, a generator. That's your problem. See Gille's answer below for details
Jul
2
comment Cryptographic Disk wiping with quantum random data?
This sounds pretty absurd. The correct way, if you are willing to incur the (usually negligible) performance hit is to encrypt data as it arrives on the disk on the fly, and decrypt it as it is read. To wipe data, destroy the 16 or 32-byte key (stored only once somewhere in the disk's firmware) using, well, any method you like (it's only a handful of bytes). You can probably come up with a clever datastructure if you need to selectively wipe data.
Jun
12
comment RSA private key d knowing e,n
No need for the quadratic sieve, the prime factors are so close that Fermat's attack actually works.
Jun
11
comment simple algorithm to encrypt/decrypt a text file
@SOJPM I suppose you are trying to keep it simple but I think it's best not to call such constructions OTP's, since they aren't.
Jun
11
comment simple algorithm to encrypt/decrypt a text file
Out of curiosity, what are the two languages in question?
Jun
7
comment Is it OK to encode data using key equal to data
@TomasM Yes. The properties of AES do not preclude that possibility (though it would probably be quite hard to exhibit a colliding pair). Just because the AES encryption is reversible with any given key does not mean that A encrypted with A cannot equal C encrypted with C (since the keys are not the same).
Jun
5
comment What's the inverse function of this decryption function?
I'm guessing the choice to ignore 0x5C bytes is a cheap mechanism for being able to pepper the ciphertext with dummy bytes to mislead people. Whether it was a calculated choice by the authors in that the bytes fgrieu mentions are never used as input (so that reliable encryption is possible) or just an "obfuscation gone wrong" horror story, no-one can say.
Jun
2
comment Euler's Totient function for semiprime numbers
Yes, that trivially follows from the definition. Are you asking whether it's "better" to compute the totient as $N - p - q + 1$ instead of $(p - 1)(q - 1)$? They are always equal, of course, since $(p - 1)(q - 1) = pq - p - q + (-1)(-1) = N - p - q + 1$.
May
30
comment Is this adaptation of one-time-pad still secure?
@yyyyyyy Don't worry, I'm sure a hundred years from now people will still be trying variations of the OTP to try and make key reuse work, even though it's provably impossible... :-)
May
26
comment Would this be a plausible authenticated encryption scheme using nested encryption?
This answer's formatting is a bit garbled and wall-of-text...
May
25
comment Would this method allow fast authenticated encryption using only a single encryption operation per block?
@Anon2000 I have removed the [flawed] tag from the title of this question (and of your other questions). There is no need to put the status of the question in its title, since it's already at the top of the question body, and the square bracket suffix notation is already used on the stackexchange network to signal duplicates/closed questions, so it's a bit confusing.
May
23
comment Factorization of a number obtained by a modular multiplication operation can reveal factors of the used operands?
"Works" for $n = 3$ as well, if the remainder of $ab$ is 2 mod 3 then one of $a$, $b$ must be 2 mod 3, and the other must be 1 mod 3. Similarly if $ab$ is 1 mod 3 then $a$ and $b$ must have the same remainder mod 3... (ignoring the trivial case where $ab$ is divisible by 3). Not sure how much useful information that leaks though.
May
11
comment Is the “Matryoshka Cipher” possible to write and implement?
You would probably have more hits by searching for "cascade cipher" or "cascading cipher", which is basically the same thing.