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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.
May
9
comment Diffie Hellman Exponentiation Implementation Problem
possible duplicate of How do institutions like banks do RSA with big primes?
May
4
comment Is ECB mode safe to use with RSA encryption?
RSA is not a block cipher and you shouldn't use it as one.
May
3
comment Why is it a bad idea to use Diffie-Hellman with a prime such that $p - 1$ is smooth?
@user24258 The order of the multiplicative group of integers modulo $p$ is $p - 1$. Loosely stated, if $p - 1$ is a product of only small factors, then you can "break down" the problem of solving DH in a group of order $p - 1$ to the problem of solving DH in a bunch of groups of way smaller order.
May
3
comment How to decrypt unusual Many Time Pad
Hint: rewrite "random[0] xor random[1] = cipher[0]" as "(all-zero plaintext) xor (random[0] xor random[1]) = cipher[0]". is this really a many-time pad? Do cipher[0] and cipher[1] actually carry any information, or are they in fact statistically independent from random[1]? Write down the probabilities on paper to make sure.
May
1
comment Block cipher does not provide security by itself
Where does the statement come from? It's meaningless with context.
Apr
30
comment Why are there no OTP ciphers?
@VincentAdvocaat We will adapt; necessity is the mother of invention, especially when it comes to protecting sensitive information. Google for "post-quantum cryptography" if you're interested in learning more!
Apr
30
comment Why are there no OTP ciphers?
@VincentAdvocaat Smaller or equal in length to. And, yes, unfortunately. But fear not, modern cryptography has evolved past the OTP, your data can be made secure in practice! Just not unconditionally :-)
Apr
30
comment Why are there no OTP ciphers?
@VincentAdvocaat I have looked at your updated question. The answer is still no, for the same reason: if your key is $k$ bits long, there is no way to encrypt more than $k$ bits of data without losing unconditional security. Encrypt even a single bit more and the ciphertext begins leaking information about the plaintext (and hence the key) and the cipher is no longer unconditionally secure.