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Feb
2
comment Generating Large Prime Numbers with Py Crypto
I think the question is off-topic because software related. The randFunc is a function that outputs random bytes. From the API docs (dlitz.net/software/pycrypto/api/current/…): getPrime(N, randfunc=None) getPrime(N:int, randfunc:callable):long Return a random N-bit prime number. If randfunc is omitted, then Random.new().read is used.
Jan
28
revised RSA-OAEP: How does it work?
add the image to the post and not as link
Jan
28
suggested approved edit on RSA-OAEP: How does it work?
Jan
20
comment What does this advantage statement signify?
Acutally it works also in the opposite sense: "how bad is..." because ad adversary that never guess correctly can easily transformed in a good guesser just inverting its answers
Jan
20
comment What does this advantage statement signify?
It really means that: you can phrase it as "how good is the adversary in guessing the right answer against a simple guessing with a random coin".
Jan
17
awarded  Autobiographer
Jan
14
comment Why gcd(r,(p-1)/r) needs to be 1 in benaloh cryptosystem
otherwise your secret key $x$ would be a $r$-residue. and the decryption would not work. I'll try to write down the whole in a complete answer during next weekend.
Jan
14
comment Why gcd(r,(p-1)/r) needs to be 1 in benaloh cryptosystem
You need those conditions to be sure that $r^2 \not | \phi(n)$
Jan
13
comment Generating a key to use with AES
Let us continue this discussion in chat.
Jan
13
comment Generating a key to use with AES
You're probably safe, but this is not the right way. The Good Thing is to verify the authenticity the ciphertext.
Jan
13
comment Generating a key to use with AES
If your problem is only to avoid someone reading your files if you forgot your USB drive at the coffee shop that would be ok.
Jan
13
comment Generating a key to use with AES
If you don't mind that an attacker could modify the ciphertext and observe the behavior of your software during the decryption (and modify one more time the ciphertext and so on), there should be no problem. If you want only to protect the confidentiality of your your files it is ok, but if you think that someone could play with your ciphertexts and your decryption function you could have a classical Oracle attack.
Jan
13
comment Generating a key to use with AES
I'm not familiar with the library you're using, but the IV is usually part of the ciphertext (the first block). The CBC (with a random IV) does not leak information about the plaintext. If you are going to use an HMAC remember that you'll need another key, you could derive it from the same password but you need to differentiate it from the encryption key. NEVER use the same key to encrypt and hmac the same message.
Jan
13
revised What is difference between meet in the middle attack and man in the middle attack?
math edit, latex notation
Jan
13
suggested approved edit on What is difference between meet in the middle attack and man in the middle attack?
Jan
13
revised Generating a key to use with AES
added 86 characters in body
Jan
13
answered Generating a key to use with AES
Jan
6
answered NIST tests for AES
Jan
5
comment Finding Private Key $d$ using RSA
Yes, you found the right algorithm. It is quite easy to implement (in python it takes only few lines)
Jan
5
revised Finding Private Key $d$ using RSA
math edit, latex notation