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 Yearling
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Oct
23
comment How bad is it to use the identity function as hash for ECDSA?
"...that your customer is stupid???" Request on mailing list doesn't imply customer. And then no one said stupid, which sounds quite harsh. But as a hint to the sad truth about cryptography: Most people don't understand it, and unless studying crypto extensively, they also have no clue what damage can be done by easy beginner's mistakes. There are countless self-proclaimed "experts", which come up with new schemes and basically re-invent things, which have been dismissed for 30 years, maybe from a different point of view, but with the same weaknesses.
Oct
23
comment Is there an additively homomorphic encryption scheme that supports calculating a square root on the ciphertext?
There is a substantial difference in using square roots over reals (with approximations, etc.) and square roots in finite rings and fields. $2^{1/2}$ mod $7$ equals both $3$ and $4$. It has nothing to do with $\sqrt{2}\approx 1.414$
Oct
23
comment Is there a format preserving cryptographically secure hash?
This is not about the construction of FPE. It is about creating a cryptographic hash function (and utilizing FPE to do so). Cryptographically secure hash functions don't protect against brute force (of small message spaces) at all. It is about collision resistance and preimage resistance. If short message spaces are a problem, you should use a construction with a key anyway (e.g. MAC).
Oct
19
comment Can it ever be impossible to invert a PRNG?
The entire argument for "impossible to uniquely invert" is not relevant for the cryptographic strength with emphasis on "uniquely": If an attacker can exclude only a portion of the possible states, he might already be able to predict the next bit with a non-negligible probability, breaking the algorithm in the cryptographic sense. Finding out the exact internal state is not required for that. As a rule of thumb: CSPRNGs and PRNGs are very different things, and the similar name is very misleading. Don't mix them up, it doesn't work.
Oct
19
comment Deciphering “easy” ciphers without hints
Well, without the actual challenge, it's quite difficult to point in the right direction. If the only "flashy" value is 81, and 3,9,27 are just average like the rest, then it is unlikely that 3,9 and 27 are correct key sizes (Factors of the correct key size don't show deviating values, just the multiples). A Vigenere cipher with a key of length 81 sounds unlikely, but not impossible. But anyway, you have found some characteristic, that means you're on the right track.
Oct
17
awarded  Yearling
Oct
16
comment Number of qubits and breaking hashes
Snake oil with traces of antibiotics doesn't make it the new wonder-drug. There is yet to find any problem, where D-Wave is actually faster than state-of-the-art classical algorithms.
Oct
16
comment Deciphering “easy” ciphers without hints
I'd suggest trying to figure out the block size first. With the aurocorrelation method you can figure that out without knowing anything about the alphabet, as long as it is not uniform distributed.
Oct
15
answered Deciphering “easy” ciphers without hints
Oct
13
comment Very short signatures? (eg: 48bits?)
I shiver, when I read the combination of "rsa" and the number "256". Pretty much anything is better than that, since such a number can be factored within seconds most likely.
Oct
13
answered How hard to solve the given mod problem
Sep
14
comment Distribution based integer factorization
The reason for that pyramid is most likely, that you choose primes with a fixed number of bits. A more common assumption is that $p<q<2p$, and in that case the most extreme possible case is $q\approx 2p$, which results in $n\approx 2p^2 \Rightarrow \sqrt{n}\approx \sqrt{2} p$, so that $\sqrt{n} / p \approx \sqrt{2}$. Thus you have an upper bound of roughly "41.4%", or if you consider the smaller prime, that's "29.3%" below. Btw, due to the well-known Fermat factorization, many RSA generators check that the numbers have a minimum distance.
Sep
7
comment How To prove Any Change to $v=a\cdot y+b$ maks $y=(a)^{−1}\cdot (v−b)$ Uni. random value
Include everything. Simple is bad, if it means vague or imprecise.
Sep
7
comment How To prove Any Change to $v=a\cdot y+b$ maks $y=(a)^{−1}\cdot (v−b)$ Uni. random value
Maybe you should clarity your setup. Who is the attacker? The server? What is the exact goal, and do you have a special security property in mind? The more clearly your statements in the question are, the more likely you get a fitting answer.
Sep
7
comment How long does it take to brute force a Wordpress password?
@BadHorsie The problem probably doesn't lie with the exchange, but with the oversight of security holes or corrupted computers (as you stated yourself, e.g. keyloggers). A recommendation here depends on what kind of website we're talking about: If it is about any kind of professional site/ business involving money, my advise is: Get an expert to analyze (and - if needed - fix) your setup. For security it's a really bad habit, that software developers think they can do anything themselves.
Sep
7
comment How long does it take to brute force a Wordpress password?
This question is really far from cryptography, and I would consider it off-topic. But to give a short answer: Being "hacked" does not necessarily imply a bad password (it will most of the time, tho). It can also be, that the implementation of security measures is simply wrong or bugged (e.g. happens when people don't follow the rule: Never implement crypto yourself), passwords were saved in the clear at the provider, and that database got stolen, ... lots of possibilities. If you want to know what happened, try to verify everything and keep speculation to a minimum.
Sep
7
comment How To prove Any Change to $v=a\cdot y+b$ maks $y=(a)^{−1}\cdot (v−b)$ Uni. random value
Well, then you have to specify what the server can and can't do. If you just say "he can change something" then he could just throw away the old number and choose a new one. You can not prove that he didn't do that. And worse, he can't prove that he didn't do that either.
Sep
7
answered How To prove Any Change to $v=a\cdot y+b$ maks $y=(a)^{−1}\cdot (v−b)$ Uni. random value
Sep
7
comment A block cipher with independent keys for each round
In the related key scenario, allowing independent round keys always offers a lot more variables to modify, but the attacker could also just use the "valid" combinations, which agree with the usual key schedule. The attack with independent round keys therefore is stronger than the common related key attack (probably even strictly stronger). And therefore "security is weaker".
Sep
7
comment A block cipher with independent keys for each round
@Dingo13 "Authors have designed a system with related (derivable) round keys by a desire of short key length ? Not ?" They key schedule is part of the overall design, and it is part of all the cryptanalysis for AES. On one hand you're speculating, and on the other this can't be answered by anyone but the designers themselves. Designing a secure block cipher (or a hash function) is one of the hardest things in cryptography, and the design features are way beyond "let's just make it nonlinear and be done with it".