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1d
comment Do most TLS 1.2 implementations express curves in a canonical form when performing EC arithmetic?
@hodgepodge Every curve is representable in Weiderstrass form, but not every curve is representable in Montgomery/Edwards form. But if it is, you can convert points from weierstrass.
1d
comment Convert ciphertexts of one encryption scheme to another without decryption
Related: Proxy re-encryption
1d
comment Caculated time One Point Multiplication with double and method
@MaartenBodewes #ADD is supposed to be the number of additions.
Feb
9
comment RSA PKCS#1, v1.5 padding output
I strongly recommend using OAEP instead of PKCS#1 v1.5 padding. Using the latter safely is very tricky and error prone.
Feb
8
comment Is there a standard for hashing a list of strings?
@pg1989 You need the first separator (the :) but not the ,. Owlstead already noted that the encoding becomes ambiguous without the : if the string starts with a number.
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)
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
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
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.
Jan
28
comment Is H'(M)=5*M + 9*H(M) mod 2^n a secure hash?
Doesn't the extended euclidean algorithm solve your supposedly NP hard problem in polynomial time?
Jan
28
comment Is H'(M)=5*M + 9*H(M) mod 2^n a secure hash?
I don't think it's safe in general - the feed-forward of your hash (based on a variant of Matyas–Meyer–Oseas) could be constructed just so that it gets cancelled out.
Jan
28
comment RSA-OAEP: How does it work?
WTF! OAEP only uses a two round Feistel network?! That's not even enough to be a random permutation (requires 3 rounds) let alone a strong random permutation (requires 4 rounds).