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comment Block cipher not resistant to collisions
@fgrieu I missed that one. How about posting it as an answer?
2d
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.
Jan
28
comment Parallel-resistant proof-of-work scheme?
Marinus Freund wrote: Admittedly, what I don't really get about puzzles that are created for the solvers, is that the solution could be transmitted instead of solved. With e.g. Bitcoin afaik no part of the protocol does create or transmit a puzzle, throw away a secret solution or anything like that. If it did, the system wouldn't work for long. This property seems always to go along with parallelism, sadly. Am I missing something?
Jan
28
revised Is H'(M)=5*M + 9*H(M) mod 2^n a secure hash?
deleted 1 character in body
Jan
28
comment Single-algorithm authenticated encryption
@RichieFrame If you use a naturally tweakable block-cipher as primitive (like Threefish) OCB is clearly just a mode-of-operation.