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10h
revised HRNG for One Time Pad
added 21 characters in body
18h
revised HRNG for One Time Pad
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19h
awarded  Nice Answer
Feb
3
comment Block cipher not resistant to collisions
@fgrieu I missed that one. How about posting it as an answer?
Feb
3
comment Block cipher not resistant to collisions
I can't think of an efficient way for finding two $x$, $y$ pairs which produce the same hash. (assuming the block-cipher is ideal and has a sufficiently large block-size)
Feb
2
revised Key Exchange MAC encryption order of events? Problem
added 10 characters in body
Feb
2
revised How does RIPEMD160 pad the message?
edited title
Feb
1
revised Computing inverses in a binary field
edited tags; edited title
Feb
1
awarded  Necromancer
Jan
31
comment A way of bruteforcing a hash value without the use of file
Why would it need to create a file containing all candidate plaintexts? It can simply generate them on the fly while iterating.
Jan
31
comment A way of bruteforcing a hash value without the use of file
Use a brute-forcing tool like hashcat or john-the-ripper.
Jan
31
comment Any stream cipher with avalanche effect?
Perhaps you want format-preserving-encryption.
Jan
31
comment Is there a standard for hashing a list of strings?
Assuming you're referring to djbs netstrings, you forgot the separator between the string and the number. Your example would be 4:john,5:smith, when using netstrings.
Jan
31
awarded  Popular Question
Jan
29
comment Is H'(M)=5*M + 9*H(M) mod 2^n a secure hash?
@MeysamGhahramani There is always exactly one solution for $M$ in $0 \le M < 2^n$. Multiplication with an odd constant is a bijection $\pmod{2^n}$.
Jan
29
comment Is H'(M)=5*M + 9*H(M) mod 2^n a secure hash?
If you want to shorten the output of $H$ why don't you simply truncate it? That's simple and secure for typical hashes.
Jan
29
comment Is H'(M)=5*M + 9*H(M) mod 2^n a secure hash?
@MeysamGhahramani I don't see how $\pmod {2^n}$ poses any problems. It just means that you need to compute $9^{-1}$ as the Modular multiplicative inverse, but that can be done efficiently using the Extended Euclidean algorithm. The inverse exists since $\operatorname{GCD}(9, 2^n)=1$.
Jan
29
comment Is this DIY remote lock protocol secure?
Your MAC $H(M+K)$ relies on the collision resistance of the hash, which is broken for MD5. $H(K+M)$ would be even worse. HMAC does not rely on collision resistance and is explicitly designed to be used as a MAC. So using HMAC is essential for such a protocol.
Jan
29
comment Is this DIY remote lock protocol secure?
Use HMAC not a silly ad-hoc MAC.
Jan
29
comment Parallel-resistant proof-of-work scheme?
@E.Rose It's a comment by Marinus (written in his now deleted answer) to you. He seems to be worried that your scheme requires a trusted party to create the puzzle, instead of using a hash as seed, like it's required by bitcoin's proof-of-work.