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8h
comment Transforming Gaussian random $[0,1] $ numbers to uniform $[0,255] $
@dylan7 A byte is just 8 bits. To get a uniformly distributed byte, take 8 uniformly distributed independent bits and put them together (as in, binary notation).
1d
comment NEW ENCODING MACHINE
I'm voting to close this question as off-topic because homework dump
2d
comment How AES treat string input to encrypt data and what will happen if the cipher key is less than that expected?
If your key has less bytes than expected then it's not really a key, just something that you're trying to pretend is a key.
Jul
25
comment Transforming Gaussian random $[0,1] $ numbers to uniform $[0,255] $
What are the parameters of the gaussian generator? What is the actual distribution of output bytes?
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
8
awarded  Enlightened
Jul
8
awarded  Nice Answer
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
21
awarded  Caucus
Jun
18
answered is this possible to use asymmetric encryption schemes as order preserving encryption
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 $(p - 1)(q - 1) = pq - p - q + (-1)(-1) = N - p - q + 1$.